Lectures on Algebraic Groups

Home , Complete variety, Irreducible component

Alexander Kleshchev

Contents

Part one: Algebraic Geometry page 1 1 General Algebra 3 2 Commutative Algebra 5 2.1 Some random facts 5 2.2 Ring extensions 8 3 Affine and Projective Algebraic Sets 18 3.1 Zariski topology 18 3.2 Nullstellensatz 20 3.3 Regular functions 22 3.4 Irreducible components 23 3.5 Category of algebraic sets 25 3.6 Products 28 3.7 Rational functions 29 3.8 Projective n-space 32 3.9 Functions 34 3.10 Product of projective algebraic sets 35 3.11 Example: Grassmann varieties and flag varieties 36 3.12 Example: Veronese variety 39 3.13 Problems 41 4 Varieties 46 4.1 Affine varieties 46 4.2 Prevarieties 49 4.3 Products 51 4.4 Varieties 53 4.5 Dimension 54 4.6 Problems 60 5 Morphisms 63 5.1 Fibers 63

iii iv Contents 5.2 Finite morphisms 64 5.3 Image of a morphism 66 5.4 Open and birational morphisms 69 5.5 Problems 70 6 Tangent spaces 72 6.1 Definition of tangent space 72 6.2 Simple points 74 6.3 Local ring of a simple point 76 6.4 Differential of a morphism 77 6.5 Module of differentials 78 6.6 Simple points revisited 83 6.7 Separable morphisms 85 6.8 Problems 87 7 Complete Varieties 88 7.1 Main Properties 88 7.2 Completeness of projective varieties 89 Part two: Algebraic Groups 91 8 Basic Concepts 93 8.1 Definition and first examples 93 8.2 First properties 95 8.3 Actions of Algebraic Groups 98 8.4 Linear Algebraic Groups 100 8.5 Problems 102 9 Lie algebra of an algebraic group 105 9.1 Definitions 105 9.2 Examples 107 9.3 Ad and ad 108 9.4 Properties of subgroups and subalgebras 110 9.5 Automorphisms and derivations 111 9.6 Problems 112 10 Quotients 114 10.1 Construction 114 10.2 Quotients 115 10.3 Normal subgroups 118 10.4 Problems 120 11 Semisimple and unipotent elements 121 11.1 Jordan-Chevalley decomposition 121 11.2 Unipotent algebraic groups 124 11.3 Problems 125 Contents v 12 Characteristic 0 theory 127 12.1 Correspondence between groups and Lie algebras 127 12.2 Semisimple groups and Lie algebras 129 12.3 Problems 130 13 Semisimple Lie algebras 131 13.1 Root systems 131 13.2 Semisimple Lie algebras 134 13.3 Construction of simple Lie algebras 137 13.4 Kostant Z-form 139 13.5 Weights and representations 140 13.6 Problems 142 14 The Chevalley construction 145 14.1 Definition and first properties 146 15 Borel subgroups and flag varieties 148 15.1 Complete varieties and Borel’s fixed point theorem 148 15.2 Borel subgroups 149 15.3 The Bruhat order 151 16 The classification of reductive algebraic groups 154 16.1 Maximal tori and the root system 154 16.2 Sketch of the classification 156 Bibliography 160

Part one Algebraic Geometry

1 General Algebra

Deﬁnition 1.0.1 A functor F : A → B is called faithful if the map

HomA(A1,A2) → HomB(F (A1),F (A2)), θ 7→ F (θ) (1.1) is injective, and F is called full if the map (1.1) is surjective.

Theorem 1.0.2 A functor F : A → B is an equivalence of categories if and only if the following two conditions hold: (i) F is full and faithful; (ii) every object of B is isomorphic to an object of the form F (A) for some A ∈ Ob A.

Proof ( ⇒ ) Let F be an equivalence of categories and G : B → A be the quasi-inverse functor. Let α : GF → idA and β : FG → idB be isomorphisms of functors. First of all, for any object B of B βB : F (G(B)) → B is an isomorphism, which gives (ii). Next, for each ϕ ∈ HomA(A1,A2) we have the commutative diagram α A-1 GF (A1) A1

GF (ϕ) ϕ ? α ? A-2 GF (A2) A2

Hence ϕ can be recovered from F (ϕ) by the formula

−1 ϕ = αA2 ◦ GF (ϕ) ◦ (αA1 ) . (1.2) This shows that F is faithful. Similarly, G is faithful. To prove that F

3 4 General Algebra is full, consider an arbitrary morphism ψ ∈ HomB(F (A1),F (A2)), and set −1 ϕ := αA2 ◦ G(ψ) ◦ (αA1 ) ∈ HomA(A1,A2).

Comparing this with (1.2) and taking into account that αA1 and αA2 are isomorphisms, we deduce that G(ψ) = GF (ϕ). As G is faithful, this implies that ψ = F (ϕ), which completes the proof that F is a full functor. ( ⇐ ) Assume that (i) and (ii) hold. In view of (i), we can (and will) identify the set HomB(F (A1),F (A2)) with the set HomA(A1,A2) for any A1,A2 ∈ Ob A. Using (ii), for each object B in B we can pick an object AB in A and an isomorphism βB : F (AB) → B. We deﬁne a functor G : B → A which will turn out to be a quasi-inverse functor to F . on the objects we set G(B) = AB for any B ∈ Ob B. To deﬁne G on the morphisms, let ψ ∈ HomB(B1,B2). G(ψ) := β−1 ◦ ψ ◦ β ∈Hom (FG(B ),FG(B )) B2 B1 B 1 2

= HomA(G(B1),G(B2)).

It is easy to see that G is a functor, and β = {βB} : FG → idB is an isomorphism of functors. Further, βF (A) = F (αA) for the unique morphism αA : GF (A) → A. Finally, it is not hard to see that α = {αA} : GF → idA is an isomorphism of functors. 2 Commutative Algebra

Here we collect some theorems from commutative algebra which are not always covered in 600 algebra. All rings and algebras are assumed to be commutative.

2.1 Some random facts

Lemma 2.1.1 Let k be a ﬁeld, f, g ∈ k[x, y], and assume that f is irreducible. If g is not divisible by f, then the system f(x, y) = g(x, y) = 0 has only ﬁnitely many solutions.

Proof See [Sh, 1.1].

Proposition 2.1.2 Let A, B be k-algebras, I / A, J / B be ideals. Then

¯ A/I ⊗k B/J → (A ⊗k B)/(A ⊗ J + I ⊗ B), a¯ ⊗ b 7→ a ⊗ b is an isomorphism of algebras.

Deﬁnition 2.1.3 A subset S of a commutative ring R is called multiplicative if 1 ∈ S and s1s2 ∈ S whenever s1, s2 ∈ S. A multiplicative subset is called proper if 0 6∈ S.

Lemma 2.1.4 Let S ⊂ R be a proper multiplicative set. Let I be an ideal of R satisfying I ∩ S = ∅. The set T of ideals J ⊇ I such that J ∩ S = ∅ has maximal elements, and each maximal element in T is a prime ideal.

5 6 Commutative Algebra Proof That the set T has maximal elements follows from Zorn Lemma. Let M be such an element. Assume that x, y ∈ R \ M. By the choice of M, M + Rx contains some s1 ∈ S and M + Ry contains some s2 ∈ S, i.e. s1 = m1 + r1x and s2 = m2 + r2y. Hence

s1s2 = (m1 + r1x)(m2 + r2y) ∈ M + Rxy. It follws that M + Rxy 6= M, i.e. xy 6∈ M.

Theorem 2.1.5 (Prime Avoidance Theorem) Let P1,...,Pn be prime ideals of the ring R. If some ideal I is contained in the union P1 ∪ · · · ∪ PN , then I is already contained in some Pi.

Proof We can assume that none of the prime ideals is contained in another, because then we could omit it. Fix an i0 ∈ {1,...,N} and for each i 6= i0 choose an fi ∈ Pi, fi 6∈ Pi0 , and choose an fi0 ∈ I, fi0 6∈ Pi0 . Q Then hi0 := fi lies in each Pi with i 6= i0 and I but not in Pi0 . Now, P hi lies in I but not in any Pi.

Lemma 2.1.6 (Nakayama’s Lemma) Let M be a ﬁnitely generated module over the ring A. Let I be an ideal in A such that for any a ∈ 1+I, aM = 0 implies M = 0. Then IM = M implies M = 0.

Proof Let m1, . . . , ml be generators of M. The condition IM = M means that l X mi = xijmj (1 ≤ i ≤ l). j=1 for some xij ∈ I. Hence

l X (xij − δij)mj = 0 (1 ≤ i ≤ l). j=1

So by Cramer’s rule, dmj = 0, where d = det(xij − δij). Hence dM = 0. But d ∈ 1 + I, so M = 0.

Corollary 2.1.7 If B ⊃ A is a ring extension, and B is ﬁnitely generated as an A-module, then IB 6= B for any proper ideal I of A.

Proof Since B contains 1, we have aB = 0 only if a = 0. Now all elements of 1 + I are non-zero for a proper ideal I, so we can apply Nakayama’s Lemma. 2.1 Some random facts 7 Corollary 2.1.8 (Nakayama’s Lemma) Let M be a ﬁnitely generated module over the ring A, M 0 ⊆ M be a submodule, and let I be an ideal in A such that all elements of 1 + I are invertible. Then IM + M 0 = M implies M 0 = M.

Proof Apply Lemma 2.1.6 to M/M 0.

Another version:

Corollary 2.1.9 (Nakayama’s Lemma) Let M be a ﬁnitely generated module over a ring A, and I be a maximal ideal of A. If IM = M, then there exists x 6∈ M such that xM = 0.

Proof Localize at I and apply Corollary 2.1.8.

Corollary 2.1.10 Let M be a ﬁnitely generated module over the ring A and let I be an ideal in A such that all elements of 1 + I are invertible. Then elements m1, . . . , mn ∈ M generate M if and only if their images generate M/IM.

0 Proof Apply Corollary 2.1.8 to M = (m1, . . . , mn).

Lemma 2.1.11 Let M be a maximal ideal of R, then the map R → RM induces the isomorphism of the ﬁelds R/M and RM /MRM . If we identify the ﬁelds via this isomorphism, then the the map R → RM also 2 2 induces the isomorphism of vector spaces M/M →˜ MRM /(MRM ) .

A field extension K/k is called separable, if either char k = 0 or char k = p > 0 and for any k-linearly independent elements x1, . . . , xn ∈ p p K, we have x1, . . . , xn are linearly independent. A filed K = k(x1, . . . , xn) is called separably generated over k if K is a finite separable extension of a purely transcendental extension of k.

Theorem 2.1.12

(i) The extension K = k(x1, . . . , xn)/k is separably generated if and only if K/k is separable. (ii) If k is perfect (in particular algebraically closed), then any field extension K/k is separable. (iii) Let F/K/k be field extensions. If F/k is separable, then K/k is separable. If F/K and K/k are separable, then F/k is separable. 8 Commutative Algebra Theorem 2.1.13 (Primitive Element Theorem) If K/k is a finite separable extension, then there is an element x ∈ K such that K = k(x).

Let L/E be a ﬁeld extension. A derivation is a map δ : E → L such that

δ(x + y) = δ(x) + δ(y) and δ(xy) = xδ(y) + δ(x)y (x, y ∈ E).

If F is a subﬁeld of E, then the derivation δ is F -derivation if it is F - linear. The space DerF (E,L) of all F -derivations is a vector space over L. With this notation, we have:

Theorem 2.1.14 (i) If E/F is separably generated then

dim DerF (E,L) = tr. degF E.

(ii) E/F is separable if and only all derivations F → L extend to derivations E → L. (iii) If char E = p > 0, then all derivations are zero on the subﬁeld Ep. In particular, if E is perfect, all derivations of E are zero.

Theorem 2.1.15 [Ma, Theorem 20.3] A regular local ring is a UFD. In particular it is an integrally closed domain.

2.2 Ring extensions

Deﬁnition 2.2.1 A ring extension of a ring R is a ring A of which R is a subring.

If A is a ring extension of R, A is a fathful R-module in a natural way. Let A be a ring extension of R and S be a subset of A. The subring of A generated by R and S is denoted R[S]. It is quite clear that R[S] consists of all R-linear combinations of products of elements of S.

Definition 2.2.2 A ring extension A of R is called finitely generated if A = R[s1, . . . , sn] for some finitely many elements s1, . . . , sn ∈ A.

The following notion resembles that of an algebraic element for ﬁeld extensions. 2.2 Ring extensions 9 Deﬁnition 2.2.3 Let A be a ring extension of R. An element α ∈ A is called integral over R if f(α) = 0 for some monic polynomial f(x) ∈ R[x]. A ring extension R ⊆ A is called integral if every element of A is integral over R.

In Proposition 2.2.5 we give two equivalent reformulations of the in- tegrality condition. For the proof we will need the following technical

Lemma 2.2.4 Let V be an R-module. Assume that v1, . . . , vn ∈ V and Pn aij ∈ R, 1 ≤ i, j ≤ n satisfy j=1 akjvj = 0 for all 1 ≤ k ≤ n. Then D := det(aij) satisﬁes Dvi = 0 for all 1 ≤ i ≤ n.

Pn Proof We expand D by the ith column to get D = k=1 akiCki, where Pn Cki is the (k, i) cofactor. We then also have k=1 akjCki = 0 for i 6= j. So n n n X X X X Dvi = akiCkivi = akiCkivi + ( akjCki)vj k=1 k=1 j6=i k=1 n n n n X X X X = akjCkivj = Cki akjvj = 0. j=1 k=1 k=1 j=1

Proposition 2.2.5 Let A be a ring extension of R and α ∈ A. The following conditions are equivalent: (i) α is integral over R. (ii) R[α] is a ﬁnitely generated R-module. (iii) There exists a faithful R[α]-module which is ﬁnitely generated as an R-module.

Proof (i) ⇒ (ii) Assume f(α) = 0 , where f(x) ∈ R[x] is monic of degree n. Let β ∈ R[α]. Then β = g(α) for some g ∈ R[x]. As f is monic, we can write g = fq+r, where deg r < n. Then β = g(α) = r(α). Thus R[α] is generated by 1, α, . . . , αn−1 as an R-module. (ii) ⇒ (iii) is clear. (iii) ⇒ (i) Let V be a faithful R[α]-module which is generated as an R-module by ﬁnitely many elements v1, . . . , vn. Write

αvi = ai1v1 + ··· + ainvn (1 ≤ i ≤ n). 10 Commutative Algebra Then

−ai1v1 − · · · − ai,i−1vi−1 + (α − aii)vi − ai,i+1vi+1 − · · · − ainvn = 0 for all 1 ≤ i ≤ n. By Lemma 2.2.4, we have Dvi = 0 for all i, where

α − a11 −a12 · · · −a1n

−a21 α − a22 · · · −a2n

D = ......

−an1 an2 ··· α − ann

As v1, . . . , vn generate V , this implies that D annihilates V . As V is faithful, D = 0. Expanding D shows that D = f(α) for some monic polynomial f(x) ∈ R[x].

Lemma 2.2.6 Let R ⊆ A ⊆ B be ring extensions. If A is finitely generated as an R-module and B is finitely generated as an A-module, then B is finitely generated as an R-module.

Proof If a1, . . . , am are generators of the R-module A and b1, . . . , bn are generators of the A-module B, then it is easy to see that {aibj} are generators of the R-module B.

Proposition 2.2.7 Let A be a ring extension of R. (i) If A is ﬁnitely generated as an R-module, then A is integral over R. (ii) If A = R[α1, . . . , αn] and α1, . . . , αn are integral over R, then A is ﬁnitely generated as an R-module and hence integral over R. (iii) If A = R[S] and every s ∈ S is integral over R, then A is integral over R.

Proof (i) Let α ∈ A. Then A is a faithful R[α]-module, and we can apply Proposition 2.2.5. (ii) Note that R[α1, . . . , αi] = R[α1, . . . , αi−1][αi]. Now apply induction, Proposition 2.2.5 and Lemma 2.2.6. (iii) Follows from (ii).

Corollary 2.2.8 Let A be a ring extension of R. The elements of A which are integral over R form a subring of A.

Proof If α1, α2 ∈ A are integral, then α1 − α2 and α1α2 belong to R[α1, α2]. So we can apply Proposition 2.2.7(ii). 2.2 Ring extensions 11 This result allows us to give the following deﬁnition

Deﬁnition 2.2.9 The integral closure of R in A ⊇ R is the ring R¯ of all elements of A that are integral over R. The ring R is integrally closed in A ⊇ R in case R¯ = R. A domain R is called integrally closed if it is integrally closed in its ﬁeld of fractions.

Example 2.2.10 The elements of the integral closure of Z in C are called algebraic integers. They form a subring of C. In fact the ﬁeld of algebraic numbers A is the quotient ﬁeld of this ring. We record some further nice properties of integral extensions.

Proposition 2.2.11 Let R, A, B be rings. (i) If R ⊆ A ⊆ B then B is integral over R if and only if B is integral over A and A is integral over R. (ii) If B is integral over A and R[B] makes sense then R[B] is integral over R[A]. (iii) If A is integral over R and ϕ : A → B is a ring homomorphism then ϕ(A) is integral over ϕ(R). (iv) If A is integral over R, then S−1A is integral over S−1R for every proper multiplicative subset S of R.

Proof (i)-(iii) is an exercise. (iv) First of all, it follows from definitions that S−1R is indeed a −1 a −1 a a 1 subring of S A. Now, let [ s ] ∈ S A. As [ s ] = [ 1 ][ s ], it suffices to a 1 −1 1 −1 show that both [ 1 ] and [ s ] are integral over S R. But s ∈ S R and a for [ 1 ] we can use the monic polynomial which annihilates a. It follows from Proposition 2.2.11(i) that the closure of R¯ in A ⊇ R is again R¯. In particular, if D is any domain and F is its field of fractions, then the closure D¯ in F is an integrally closed domain (since the quotient field of D¯ is also F ). We recall that a domain R is called a unique factorization domain or UFD if every non-zero non-unit element of R can be written as a product of irreducible elements, which is unique up to a permutation and units.

Proposition 2.2.12 Every UFD is integrally closed.

a Proof Let R be a UFD and F be its ﬁeld of fractions. Let b ∈ F be integral over R. We may assume that no irreducible element of R divides 12 Commutative Algebra

n n−1 both a and b. There is a monic polynomial f(x) = x +rn−1x +···+ a n n−1 n r0 ∈ R[x] with f( b ) = 0, which implies a + rn−1a b + ··· + r0b = 0. So, if p ∈ R is an irreducible element dividing b then p divides an, and a hence p divides a, a contradiction. Therefore b is a unit and b ∈ R.

Proposition 2.2.13 If a domain R is integrally closed, then so is S−1R for any proper multiplicative subset S of R.

Proof Exercise.

Example 2.2.14 The ring Z[i] of Gaussian integers is Euclidean (the degree function is ∂(a + bi) = a2 + b2, hence it is a UFD, and so it is integrally closed by Proposition 2.2.13. On the other hand consider the ring Z[2i]. The quotient ﬁeld of both Z[i] and Z[2i] is Q(i), and we have Z[2i] ⊂ Z[i] ⊂ Q[i]. Clearly Z[2i] is not integrally closed, as i 6∈ Z[2i] is integral over it. It is easy to see that Z[2i] = Z[i].

Theorem 2.2.15 If R is integrally closed, then so is R[x1, . . . , xr].

Next we are going to address the question of how prime ideals of R and A are related if A ⊇ R is an integral extension.

Deﬁnition 2.2.16 Let R ⊆ A be a ring extension. We say that a prime ideal P of A lies over a prime ideal p of R if P ∩ R = p.

The following lemma is a key technical trick.

Lemma 2.2.17 Let A ⊇ R be an integral ring extension, p be a prime ideal of R, and S := R \ p. (i) Let I be an ideal of A avoiding S, and P be an ideal of A maximal among the ideals of A which contain I and avoid S. Then P is a prime ideal of A lying over p. (ii) If P is a prime ideal of A which lies over p, then P is maximal in the set T of all ideals in A which avoid S.

Proof (i) Clearly, S is a proper multiplicative subset of A. So P is prime in view of Lemma 2.1.4. We claim that P ∩ R = p. That P ∩ R ⊆ p is clear as P ∩ S = ∅. Assume that P ∩ R ( p. Let c ∈ p \ P . By the maximality of P , p + αc = s ∈ S for some p ∈ P and α ∈ A. As A is integral over R, we 2.2 Ring extensions 13 have

n n−1 0 = α + rn−1α + ··· + r0

n for some r0, . . . , rn−1 ∈ R. Multiplying by c yields

n n n−1 n−1 n 0 = c α + crn−1c α + ··· + c r0 n n−1 n = (s − p) + crn−1(s − p) + ··· + c r0.

If we decompose the last expression as the sum of monomials, then the part which does not involve any positive powers of p looks like

n n−1 n x := s + crn−1s + ··· + c r0.

It follows that x ∈ P . On the other hand, x ∈ R, so x ∈ R ∩ P ⊆ p. Now c ∈ p implies sn ∈ p. As p is prime, s ∈ p, a contradiction. (ii) If P is not maximal in T , then there exists an ideal I in T which properly contains P . As I still avoids S, it also lies over p. Take u ∈ I\P . Then u 6∈ R and u is integral over R. So the set of all polynomials f ∈ R[x] such that deg f ≥ 1 and f(u) ∈ P is non-empty. Take such Pn i f(x) = i=0 rix of minimal possible degree. We have

n n−1 u + rn−1u + ··· + r0 ∈ P ⊆ I, whence r0 ∈ R ∩ I = p = R ∩ P ⊆ P . Therefore

n n−1 n−1 n−2 u + rn−1u + ··· + r1u = u(u + rn−1u + ··· + r1) ∈ P.

n−1 n−2 By the choice of u and minimality of deg f, u 6∈ P and u +rn−1u + ··· + r1 6∈ P . We have contradiction because P is prime.

Corollary 2.2.18 (Lying Over Theorem) If A is integral over P then for every prime ideal p of R there exists a prime ideal P of A which lies over p. More generally, for every ideal I of A such that I ∩ R ⊆ p there exists a prime ideal P of A which contains I and lies over p.

Corollary 2.2.19 (Going Up Theorem) Let A ⊇ R be an integral ring extension, and p1 ⊆ p2 be prime ideals in R. If P1 is a prime ideal of A lying over p1, then there exists a prime ideal P2 of A such that P1 ⊆ P2 and P2 lies over p2.

Proof Take p = p2 and I = P1 in Lemma 2.2.17(i). 14 Commutative Algebra Corollary 2.2.20 (Incomparability) Let A ⊇ R be an integral ring extension, and P1,P2 be prime ideals of A lying over a prime ideal p of R. Then P1 ⊆ P2 implies P1 = P2.

Proof Use Lemma 2.2.17(ii). The relation between prime ideals established above has further nice properties.

Theorem 2.2.21 (Maximality) Let A ⊇ R be an integral ring extension, and P be a prime ideal of A lying over a prime ideal p of R. Then P is maximal if and only if p is maximal.

Proof If p is not maximal, we can ﬁnd a maximal ideal m ) p. By the Going Up Theorem, there is an ideal M of A lying over m and containing P . It is clear that M actually containg P properly, and so P is not maximal. Conversely, let p be maximal in R. Let M be a maximal ideal containing P . Then M ∩ R ⊇ P ∩ R = p and we cannot have M ∩ R = R, as 1R = 1S 6∈ M. It follows that M ∩ R = p. Now M = P by Incompa- rability Theorem. The previous results can be used to prove some useful properties concerning extensions of homomorphisms.

Lemma 2.2.22 Let A ⊇ R be an integral ring extension. If R is a ﬁeld then A ⊇ R is an algebraic ﬁeld extension.

Proof Let α ∈ A be a non-zero element. Then α is algebraic over R, hence R[α] ⊆ A is a ﬁeld, and α is invertible. Hence A is a ﬁeld.

Proposition 2.2.23 Let A be integral over R. Every homomorphism ϕ of R to an algebraically closed ﬁeld F can be extended to A.

Proof If R is a field, then A is an algebraic field extension of R by Lemma 2.2.22. Now the result follows from Proposition ??. If R is local, then ker ϕ is the maximal ideal m of R. By Lying Over and Maximality Theorems, there is an ideal M of A lying over m. The inclusion R → A then induces an embedding of fields R/m → A/M, which we use to identify R/m with a subfield of A/M. Note that the field extension A/M ⊇ R/m is algebraic. Since ker ϕ = m, ϕ factors 2.2 Ring extensions 15 through the projection R → R/m. The resulting homomorphism ϕ : R/m → F can be extended to ψ : A/M → F by Proposition ??. Now if π : A → A/M is the natural projection, then ψ ◦ π is the desired extension of ϕ. Now we consider the general case. Let p := ker ϕ, a prime ideal in R, and S = R \ p. Then S−1A is integral over S−1R by Proposi- −1 tion 2.2.11(iv). Now S R = Rp is local. By the universal property of localizations, ϕ extends to a ring homomorphismϕ ˆ : S−1R → F . By the local case,ϕ ˆ extends to ψˆ : S−1A → F , and the desired extension ψ : A → F is obtained by composing ψ with the natural homomorphism A → S−1A.

Proposition 2.2.24 Every homomorphism of a field k into an algebraically closed field can be extended to every finitely generated ring extension of k.

Proof Let ϕ : k → F be a homomorphism to an algebraically closed field F and R be a finitely generated ring extension of k, so that R = k[α1, . . . , αn] for some α1, . . . , αn ∈ R. First assume that R is a field. By Proposition 2.2.23, we may assume that R is not algebraic over k. Let {β1, . . . , βt} be a (necessarily finite) transcendence base of R over k. Each α ∈ R is algebraic over k k(β1, . . . , βt), i.e. satisfies a polynomial akα + ··· + a1α + a0 = 0 with coefficients ak, . . . , a0 ∈ k(β1, . . . , βt), ak 6= 0. Multiplying by a common denominator yields a polynomial equation

k bkα + ··· + b1α + b0 = 0 with coeﬃcients bk, . . . , b0 ∈ k[β1, . . . , βt], bk 6= 0. Hence α is in- 1 tegral over k[β1, . . . , βt, ]. Applying this to α1, . . . , αn yields non- bk zero c1, . . . , cn ∈ k[β1, . . . , βt] such that α1, . . . , αn are integral over 1 1 k[β1, . . . , βt, ,..., ]. Set c = c1 . . . cn. Then α1, . . . , αn are inte- c1 cn 1 1 gral over k[β1, . . . , βt, c ], and hence R is integral over k[β1, . . . , βt, c ], see Proposition 2.2.7(ii). Let cϕ be the image of c under the homomorphism ∼ k[β1, . . . , βt] = k[x1, . . . , xt] → F [x1, . . . , xt] induced by ϕ. As F is inﬁnite there exist γ1, . . . , γt ∈ F such that ϕ c (γ1, . . . , γt) 6= 0. By the universal property of polynomial rings, there exists a homomorphism ψ : k[β1, . . . , βt] → F which extends ϕ and sends 16 Commutative Algebra

β1, . . . , βt to γ1, . . . , γt, respectively. The universal property of localiza- 1 tions yields an extension of ψ to ring k[β1, . . . , βt, c ] = k[β1, . . . , βt]c. Now Proposition 2.2.23 extends ϕ to R, which completes the case where R is a ﬁeld.

Now, let R = k[α1, . . . , αn] be any finitely generated ring extension of k. Let m be a maximal ideal of R and π : R → R/m be the natural projection. Then R/m is a field extension of π(k) =∼ k generated by π(α1), . . . , π(αn). By the first part of the proof, every homomorphism of π(k) into F extends to R/m. Therefore every homomorphism of k =∼ π(k) extends to R.

Let K/k be a finite field extension. Consider K as a k-vector space. Then the map x 7→ ax is a k-linear map of this vector space. Define the × norm NK/k(a) to be the determinant of this map. Note that NK/k|K : K× → k× is a group homomorphism.

Lemma 2.2.25 If a = a1, . . . , as be the roots with multiplicity of the minimal polynomial irr (a, k) (in some extension of the ﬁeld K), then Qs [K:k(a)] NK/k(a) = ( i=1 ai) .

i Proof If 1 = v1, v2, . . . , vr is a basis of K over k(u), then {a vj | 0 ≤ i < s, 1 ≤ j ≤ r} is a basis of K over k in which the matrix of the map x 7→ ax is block diagonal with blocks all equal to the companion matrix of irr (a, k).

Lemma 2.2.26 Let S ⊆ R be integral domains with ﬁelds of fractions k ⊆ K, S be integrally closed, and r ∈ R be integral over S. Then irr (r, k) ∈ S[x].

Proof Let F be an extension of K which contains all roots r = r1, . . . , rs of irr (r, k). Then each ri is integral over S. So the coeﬃcients of irr (r, k), being polynomials in the ri are also integral over S. As S is integrally closed, the coeﬃcients belong to S.

Corollary 2.2.27 Let S ⊆ R be domains with fields of fractions k ⊆ K such that the field extension K/k is finite. Assume that the ring extension S ⊆ R is integral and that S is integrally closed. Then NK/k(r) ∈ S for any r ∈ R.

Proof Apply Lemmas 2.2.25 and 2.2.26. 2.2 Ring extensions 17 Lemma 2.2.28 (Noether’s Normalization Lemma) Let k be a field, and R = k[x1, . . . , xn] be a domain, finitely generated over k with the field of fractions F . If tr. degk F = d, then there exist algebraically independent over k elements S1,...,Sd ∈ R such that R is integral over k[S1,...,Sd].

Theorem 2.2.29 (Going Down Theorem) Let S ⊆ R be an integral ring extension and S be integrally closed. Let P1 ⊇ P2 be prime ideals of S, and Q1 be a prime ideal of R lying over P1. Then there exists a prime ideal Q2 ⊆ Q1 lying over P2. 3 Aﬃne and Projective Algebraic Sets

3.1 Zariski topology Algebraic geometry is the subject which studies (algebraic) varieties. Naively, varieties are just algebraic sets. Throughout we fix an algebraically closed ground field k. (It is much harder to develop algebraic geometry over non-algebraically closed fields and we will not try to do this). Denote by An the affine space kn—this is just the set of all n-tuples of elements of k.

Deﬁnition 3.1.1 Let S ⊂ k[T1,...,Tn]. A zero of the set S is an n element (x1, . . . , xn) of A such that f(x1, . . . , xn) = 0 for all f ∈ S. The zero set of S is the set Z(S) of all zeros of S. An algebraic set in n A (or aﬃne algebraic set) is the zero set of some set S ⊂ k[T1,...,Tn], in which case S is called a set of equations of the algebraic set.

Example 3.1.2 The straight line x+y −1 = 0 and the ‘circle’ x2 +y2 − 1 = 0 are examples of algebraic sets in C2. More generally, algebraic sets in C2 with a single equation are called complex algebraic curves. Note that the curve given by the equation (x + y − 1)(x2 + y2 − 1) = 0 is the union of the line and the ‘circle’ above. On the other hand, the zero set of {x + y − 4, x2 + y2 − 1} consists of two points (1, 0) and (0, 1). Finally, two more examples: ∅ = Z(1), and C2 = Z(0).

Note that Z(S) = Z((S)), where (S) is the ideal of k[T1,...,Tn] generated by S. Therefore every algebraic set is the zero set of some ideal. Since k[T1,...,Tn] is noetherian by Hilbert’s Basis Theorem, every algebraic set is the sero set of a ﬁnite set of polynomials.

Example 3.1.3 Let us try to ‘classify’ algebraic sets in A1 and A2.

18 3.1 Zariski topology 19

(i) Algebraic sets in A1 are A1 itself and all finite subsets (including ∅). (ii) Let X be an algebraic set in A2. It is given by a system of polynomial equations: f1(T1,T2) = ··· = fm(T1,T2) = 0. If all 2 polynomials are zero, we get X = A . If f1, . . . , fm do not have a common divisor, then our system has only finitely many solutions, see Lemma 2.1.1. Finally, let all fi have greatest common divisor d(T1,T2). Then fi = dgi, where the polynomials gi(T1,T2) do not have a common divisor. Now, X = X1 ∪ X2, where X1 is given by the system g1 = ··· = gm = 0, and X2 is given by d = 0. As above X1 is a finite (possibly empty) set of points, while X2 is given by one non-trivial equation d = 0 (and can be thought of as a ‘curve’ in A2).

Proposition 3.1.4

(i) Every intersection of algebraic sets is an algebraic set; the union of ﬁnitely many algebraic sets is an algebraic set. (ii) An and ∅ are algebraic sets in An.

Proof (i) Let (Xj = Z(Ij))j∈J be a family of algebraic sets, given as zero sets of certain ideals Ij. To see that their intersection is again an P algebraic set, it is enough to note that ∩j∈J Z(Ij) = Z( j∈J Ij). For the union, let Z(I) and Z(J) be algebraic sets corresponding to ideals I and J, and note that Z(I) ∪ Z(J) = Z(I ∩ J) (why?). (ii) An = Z(0) and ∅ = Z(1).

The proposition above shows that algebraic sets in An are closed sets of some topology. This topology is called the Zariski topology. Zariski toplology on An also induces Zariski topology on any subset of An, in particular algebraic set. This topology is very weird and it takes time to get used to it. The main unintuitive thing here is that the topology is ‘highly non-Hausdorf’—its open sets are huge. For example, we saw above that proper closed sets in k are exactly the finite sets, and so any two non-empty open sets intersect non-trivially. n Let f ∈ k[T1,...,Tn]. The corresponding principal open set is A \ Z(f) = {x ∈ An | f(x) 6= 0}. It is easy to see that each open set in An is a finite union of principal open sets, so principal open sets form a base of Zariski topology. 20 Affine and Projective Algebraic Sets 3.2 Nullstellensatz The most important theorem of algebraic geometry is called Hilbert’s Nullstellensatz (or theorem on zeros). It has many equivalent reformulations and many corollaries. The idea of the theorem is to relate algebraic n sets in A (geometry) and ideals in k[T1,...,Tn] (commutative algebra). We have two obvious maps

n Z : {ideals in k[T1,...,Tn]} → {algebraic sets in A } and n I : {algebraic sets in A } → {ideals in k[T1,...,Tn]}.

We have already deﬁned Z(J) for an ideal J in k[T1,...,Tn]. As for I, let X be any subset of An. Then the ideal I(X) is deﬁned to be

I(X) := {f ∈ k[T1,...,Tn] | f(x1, . . . , xn) = 0 for all (x1, . . . , xn) ∈ X}.

Lemma 3.2.1 Let X be any subset of An. Then Z(I(X)) = X¯, the closure of X in Zariski topology. In particular, if X is an algebraic set, then Z(I(X)) = X.

Proof We have to show that for any algebraic set Z(J) containing X we actually have Z(I(X)) ⊆ Z(J). Well, as X ⊆ Z(J), we have I(X) ⊇ J, which in turn implies Z(I(X)) ⊆ Z(J). Note, however, that Z and I do not give us a one-to one correspondence. For example, in A1 we have Z((T )) = Z((T 2)) = {0}, that is the diﬀerent ideals (T ) and (T 2) give the same algebraic set. Also, note that I({0}) = (T ) 6= (T 2). Nullstellensatz sorts out problems like this in a very satisfactory way. The ﬁrst formulation of the Nulltellensatz is as follows (don’t forget that k is algebraically closed, otherwise the theorem is wrong):

Theorem 3.2.2 (Hilbert’s Nullstellensatz)√ Let J be an ideal of k[T1,...,Tn]. Then I(Z(J)) = J. √ Proof√ First of all, it is easy to see that J ⊆ I(Z(J)). Indeed, let f ∈ J. Then f n ∈ J. Then f n is zero at every point of Z(J). But this implies that f is zero at every point of Z(J), i.e. f ∈ I(Z(J)). The converse is much deeper. Let f ∈ I(Z(J)) and assume that no power of f belongs to J. Applying Lemma 2.1.4 to the multiplicative set {1, f, f 2,... } yields a prime ideal P containing J but not f. Let 3.2 Nullstellensatz 21

R = k[T1,...,Tn]/P and π : k[T1,...,Tn] → R be the natural projection. Then R is a domain which is generated over π(k) =∼ k by α1 := π(T1), . . . , αn := π(Tn). We identify k and π(k), and so π can be considered as a homomorphism of k-algebras. Under this agreement, y := f(α1, . . . , αn) = π(f) 6= 0, non-zero element of R, as f 6∈ P . By Proposition 2.2.24, the identity isomorphism k → k can be ex- 1 tended to a homomorphism ψ from the subring k[α1, . . . , αn, y ] of the fraction ﬁeld of R to k. Then ψ(y) 6= 0. So

f(ψ(α1), . . . , ψ(αn)) = ψ(f(α1, . . . , αn)) = ψ(y) 6= 0.

On the other hand, for any g ∈ J ⊆ P we have

g(ψ(α1), . . . , ψ(αn)) = ψ(g(α1, . . . , αn)) = ψ(g(π(T1), . . . , π(Tn)))

= ψ(π(g(T1,...,Tn))) = ψ(π(g)) = ψ(0) = 0.

Thus (ψ(α1), . . . , ψ(αn)) is a zero of J but not of f, i.e. f 6∈ I(Z(J)), a contradiction.

Deﬁnition√ 3.2.3 We say that an ideal I of a commutative ring R is radical if I = I.

The following corollary is also often called Nullstellensatz.

Corollary 3.2.4 The maps I and Z induce an order-reversing bijection n between algebraic sets in A and radical ideals in k[T1,...,Tn].

Proof Note that I(X) is always a radical ideal for any subset X ⊆ An. Now the result follows from Theorem 3.2.2 and Lemma 3.2.1.

Corollary 3.2.5 Let J and J be two ideals of k[T ,...,T ]. Then 1 √ 2 √ 1 n Z(J1) = Z(J2) if and only if J1 = J2. √ Proof It is clear that Z(J) = Z( J) for any ideal J, which gives the ‘if’-part. The converse follows from Theorem 3.2.2.

Corollary 3.2.6 Every proper ideal of k[T1,...,Tn] has at least one zero in An. √ Proof If I = k[T1,...,Tn], then I = k[T1,...,Tn]. Now the result follows from above. 22 Aﬃne and Projective Algebraic Sets

n Let x = (x1, . . . , xn) ∈ A . Denote I({x}) by Mx, i.e.

Mx = {f ∈ k[T1,...,Tn] | f(x1, . . . , xn) = 0}.

Corollary 3.2.7 The mapping x 7→ Mx is a one-to-one correspondence n between A and the maximal ideals of k[T1,...,Tn].

Proof Note that the maximal ideals are radical and apply Nullstellen- satz.

3.3 Regular functions n Let X ⊆ A be an algebraic set. Every polynomial f ∈ k[T1,...,Tn] deﬁnes a k-valued function on An and hence on X via restriction. Such functions are called regular functions on X. The regular functions form a k-algebra with respect to the obvious ‘point-wise operations’. The algebra is called the coordinate algebra (or coordinate ring) of X (or simply the algebra/ring of regular functions on X) and denoted k[X]. Clearly, ∼ k[X] = k[T1,...,Tn]/I(X). If I is an ideal of k[X] then we write Z(I) for the set of all points x ∈ X such that f(x) = 0 for every f ∈ I, and if Z is a subset of X we denote by I(Z) the ideal of k[X] which consists of all functions f ∈ k[X] such that f(z) = 0 for every z ∈ Z. Note that closed subsets of X all look like Z(I). Now the Nullstellensatz and the correspondence theorem for ideals imply:

Theorem 3.3.1 (Hilbert’s Nullstellensatz) Let X be an algebraic set. √ (i) If J is an ideal of k[X], then I(Z(J)) = J. (ii) The maps I and Z induce an order-reversing bijection between closed sets in X and radical ideals in k[X]. (iii) Every proper ideal of k[X] has at least one zero in X.

(iv) The mapping x 7→ Mx = {f ∈ k[X] | f(x) = 0} is a one-to-one correspondence between X and the maximal ideals of k[X].

Definition 3.3.2 A commutative finitely generated k-algebra without nilpotent elements is called an affine k-algebra. 3.4 Irreducible components 23 Proposition 3.3.3 (i) Let X be an algebraic set. Then k[X] is an affine k-algebra. (ii) Every affine k-algebra A is isomorphic to k[X] for some affine algebraic set X.

Proof (i) clear. For (ii), if A = k[α1, . . . , αn] is an k-algebra generated ∼ by α1, . . . , αn, then by the universal property of polynomial rings, A = k[T1,...,Tn]/I for some radical ideal I. So I = I(X) for some algebraic set X by the Nulltellensatz.

Let f ∈ k[X]. The corresponding principal open set is

Xf := X \ Z(f) = {x ∈ X | f(x) 6= 0}. (3.1) Each open set in X is a ﬁnite union of principal open sets, so principal open sets form a base of Zariski topology.

Example 3.3.4 (i) If X is a point, then k[X] = k. n (ii) If X = A , then k[X] = k[T1,...,Tn]. 2 (iii) Let X ⊂ A be given by the equation T1T2 = 1. Then k[X] is ∼ −1 isomorphic to the localization k[t]t = k[t, t ].

3.4 Irreducible components

Deﬁnition 3.4.1 A topological space is noetherian if its open sets satisfy the ascending chain condition. A topological space is irreducible if it cannot be written as a union of its two proper closed subsets.

Note that a non-empty open subset of an irreducible topological space X is dense in X, and that any two non-empty open subsets of X intersect non-trivially. Problem 3.13.18 contains some further important properties of irreducible spaces.

Lemma 3.4.2 An with Zariski topology is noetherian. Hence the same is true for any subspace of An.

Proof An ascending chain of open sets corresponds to a descending chain of closed sets, which, by the Nullstellensatz, corresponds to an 24 Aﬃne and Projective Algebraic Sets ascending chain of radical ideals of k[T1,...,Tn], which stabilizes since k[T1,...,Tn] is noetherian.

Lemma 3.4.3 Algebraic set X ⊆ An is irreducible if and only if the ideal I(X) is prime.

Proof If X is irreducible and f1, f2 ∈ k[T1,...,Tn] with f1f2 ∈ I(X), then X ⊆ Z((f1)) ∪ Z((f2)), and we deduce that X ⊆ Z((f1)) or X ⊆ Z((f2)), i.e. f1 ∈ I(X) or f2 ∈ I(X). Conversely, if I(X) is prime and X = X1 ∪ X2 for proper closed subsets X1,X2, then there are polynomials fi ∈ I(Xi) with fi 6∈ I(X). But f1f2 ∈ I(X), contradiction.

Since prime ideals are radical, Lemma 3.4.3 allows us to further re- fine the one-to-one correspondence between radical ideals and algebraic sets: under this correspondence prime ideals correspond to irreducible algebraic sets. Also note that X is irreducible if and only if k[X] is a domain. So for irreducible algebraic sets X we can form the quotient field of k[X] is called the field of rational functions on X and denoted k(X). In a natural way, k(X) is a field extension of k. We now establish a general fact on noetherian topological spaces, which in some sense reduces the study of algebraic sets to that of irreducible algebraic sets.

Proposition 3.4.4 Let X be a noetherian topological space. Then X is a ﬁnite union X = X1 ∪ · · · ∪ Xr of irreducible closed subsets. If one assumes that Xi 6⊆ Xj for all i 6= j then the Xi are unique up to permutation. They are called the irreducible components of X and can be characterized as the maximal irreducible closed subsets of X.

Proof Let X be a topological noetherian space for which the first state- 0 ment is false. Then X is reducible, hence X = X1 ∪ X1 for proper 0 closed subsets X1,X1. Moreover, the first statement is false for at 0 least one of X1,X1. Continuing this way, we get an infinite chain X ) X1 ) ··· ) X2 ) ... of closed subsets, which is a contradiction, as X is noetherian. To show uniqueness, assume that we have two irredundant decom- 0 0 positions X = X1 ∪ · · · ∪ Xr and X = X1 ∪ · · · ∪ Xs. For each i, 0 0 Xi ⊆ (X1 ∩ Xi) ∪ · · · ∪ (Xs ∩ Xi), so by irreducibility of Xi we may as- 0 0 sume that Xi ⊆ Xσ(i) for some σ(i). For the same reason, Xj ⊆ Xτ(j) for 3.5 Category of algebraic sets 25 some τ(j). Now the irredundancy of the decompositions implies that σ and τ are mutually inverse bijections between {1, . . . , r} and {1, . . . , s}, 0 and Xi = Xσ(i) for all i.

3.5 Category of algebraic sets We now deﬁne morphisms between algebraic sets. Let X ⊆ An, Y ⊆ Am be two algebraic sets and consider a map ϕ : X → Y . Let T1,...,Tn n m and S1,...,Sm be the coordinate functions on A and A , respectively. Denote Si ◦ ϕ by ϕi for all 1 ≤ i ≤ m. So that we can think of ϕ as the m-tuple of functions ϕ = (ϕ1, . . . , ϕm), where ϕi : X → k, and m ϕ(x) = (ϕ1(x), . . . , ϕm(x)) ∈ A . The map ϕ : X → Y is called a morphism of algebraic sets or a regular map from X to Y if each function ϕi : X → k, 1 ≤ i ≤ n is a regular function on X. It is easy to see that algebraic sets and regular maps form a category, in particular a composition of regular maps is a regular map again. Now, let ϕ : X → Y be a morphism of algebraic sets as above. This morphism deﬁnes the ‘dual’ morphism ϕ∗ : k[Y ] → k[X] of coordinate algebras, as follows:

ϕ∗ : k[Y ] → k[X]: f 7→ f ◦ ϕ.

It is clear that ϕ∗ is a homomorphism of k-algebras. Moreover, (ϕ◦ψ)∗ = ψ∗ ◦ ϕ∗ and id∗ = id, i.e. we have a contravariant functor F from the category of algebraic sets to the category of aﬃne k-algebras. To reiterate: F(X) = k[X] and F(ϕ) = ϕ∗.

Theorem 3.5.1 The functor F from the category of algebraic sets (over k) to the category of aﬃne k-algebras is a (contravariant) equivalence of categories.

Proof In view of Theorem 1.0.2 (for contravariant functors) and Propo- sition 3.3.3(ii) we just need to show that ϕ 7→ ϕ∗ establishes a one-to one correspondence between regular maps ϕ : X → Y and algebra homomorphisms k[Y ] → k[X], for arbitrary ﬁxed algebraic sets X ⊆ An m and Y ⊆ A . Let T1,...,Tn and S1,...,Sm be the coordinate functions on An and Am, respectively.

Let α : k[Y ] → k[X] be an k-algebra homomorphism. Set sj := Sj|Y ∈ k[Y ], 1 ≤ j ≤ m. Then α(sj) are regular functions on X. Deﬁne 26 Aﬃne and Projective Algebraic Sets

m the regular map α∗ : X → A as follows:

α∗ := (α(s1), . . . , α(sm)).

We claim that in fact α∗(X) ⊆ Y . Indeed, let x ∈ X and f = P k1 km k ckS1 ...Sm ∈ I(Y ), where k stands for the m-tuple (k1, . . . , km). It suﬃces to prove that f(α∗(x)) = 0. Using f(s1, . . . , sm) = 0 and the fact that α is an algebra homomorphism, we have

f(α∗(x)) = f(α(s1)(x), . . . , α(sm)(x))

X k1 km = ck(α(s1)(x)) ... (α(sm)(x)) k

X k1 km = α( cks1 . . . sm )(x) k

= α(f(s1, . . . , sm))(x) = 0. Now, to complete the proof of the theorem, it suﬃces to check that ∗ ∗ (ϕ )∗ = ϕ and (α∗) = α for any regular map ϕ : X → Y and any k-algebra homomorphism α : k[Y ] → k[X]. Well,

∗ ∗ ∗ (ϕ )∗ = (ϕ (s1), . . . , ϕ (sm)) = (ϕ1, . . . , ϕm) = ϕ. On the other hand,

∗ ((α∗) )(si) = si ◦ α∗ = α(si)

∗ for any 1 ≤ i ≤ m. Since the si generate k[Y ], this implies that (α∗) = α.

Corollary 3.5.2 Two (aﬃne) algebraic sets are isomorphic if and only if their coordinate algebras are isomorphic.

Lemma 3.5.3 Regular maps are continuous in the Zariski topology.

Proof Let ϕ : X → Y ⊆ Am be a regular map. As the topology on Y is induced by that on Am, it suﬃces to prove that any regular map ϕ : X → Am is continuous. Let Z = Z(I) be a closed subset of Am. We claim that ϕ−1(Z) = Z(J) where J is the ideal of k[X] generated by ϕ∗(I). Well, if x ∈ Z(J), then f(ϕ(x)) = ϕ∗(f)(x) = 0 for any f ∈ I, so ϕ(x) ∈ Z(I), i.e. x ∈ ϕ−1(Z). The argument is easily reversed.

Remark 3.5.4 Note that regular maps from X to Y usually do not exhaust all continuous maps from X to Y , so the category of algebraic 3.5 Category of algebraic sets 27 sets is not a full subcategory of the category of topological spaces. For example, if X = Y = C, the closed subsets in X and Y are exactly the finite subsets, and there are lots of non-polynomial maps from C to C such that inverse image of a finite subset is finite (describe one!).

Remark 3.5.5 The proof of Proposition 3.3.3 allows us to ‘find’ X from k[X]. More careful look at the proof however shows that we do not have a functor from affine algebras to algebraic sets, as ‘recovering’ X from k[X] is not canonical—it depends on the choice of generators in k[X], so only ‘recover X up to isomorphism’. The problem here is that our definition of algebraic sets is not a ‘right one’—it relies on embedding into some An, and this is something which we want to eventually avoid. At this stage, we can at least canonically recover X from k[X] as a topological space. Indeed, we know that as a set, X is in bijection with the set Specm k[X] of maximal ideals of the algebra k[X]. So if we want to construct a reasonable quasi-inverse functor G to the functor F, we could associate Specm k[X] to k[X]. Now make Specm k[X] into a topological space by considering the topology whose basis consists of all Xf := {M ∈ Specm | f 6∈ M}. Then x 7→ Mx is a homeomorphism from X to Specm X. Finally, if α : k[Y ] → k[X] is an algebra homomorphism define G(α) : Specm k[X] → Specm k[Y ] as follows: if M ∈ Specm k[X] then G(M) is the maximal ideal N in k[Y ] containing α−1(M). Note that if we identify X with Specm k[X] as above, and ϕ : X → Y is a ∗ morphism, then ϕ = G(ϕ )—in other words, Mϕ(x) is the maximal ideal ∗ −1 of k[Y ] containing (ϕ ) (Mx).

Example 3.5.6 (i) The notion of a regular function on X and a regular map from X to k coincide.

(ii) Projection f(T1,T2) = T1 is a regular map of the curve T1T2 = 1 to k. (iii) The map f(t) = (t, tk) is an isomorphism from k to the curve y = xk. (iv) The map α(t) = (t2, t3) is a regular map from k to the curve X ⊂ A2 given by x3 = y2. This map is clearly one-to-one, but it is not an isomorphism (even though it is a homeomorphism!) Indeed, any regular function on X has a representative p(x) + q(x)y in k[x, y] for some p, q ∈ k[x]. Now α∗(p(x)+q(x)y) = p(t2)+q(t2)t3, which is never equal to t, for example. So α∗ is not surjective. 28 Aﬃne and Projective Algebraic Sets

Moreover, one can see that X is not isomorphic to A1, since k[X] =6∼ k[T ].

Example 3.5.7 Let X be an algebraic set, and G be its finite group of automorphisms. Then G is also a group of automorphisms of the algebra A = k[X]. Suppose that char k6 | |G|. Then the invariant algebra AG is an affine algebra (the only non-trivial thing here is that it is finitely generated, which can be looked up in [Sh, Appendix].) So there is an algebraic set Y with k[Y ] = AG, and the regular map π : X → Y with π∗ being the embedding of AG into A. This algebraic set Y is called the quotient of X by G and is denoted X/G. The map π leads to a natural one-to-one correspondence between the elements of X/G and the G-orbits on X.

Indeed, we claim that for x1, x2 ∈ X, one has π(v1) = π(v2) if and only if x1 and x2 are in the same G-orbit. Well, if x2 = gx1, then G f(x1) = f(x2) for all f ∈ A = k[Y ], and so π(x1) = π(x2). Conversely, if x1 and x2 are not in the same orbit, then let f ∈ k[X] be a function with f(gT2) = 1 and f(gT1) = 0 for all g ∈ G (why does it exist?). 1 P ∗ G Then the average function S(f) := |G| g∈G g f belongs to A and ‘separates’ x1 from x2. So π(x1) 6= π(x2). Finally, in view of Remark 3.5.5, the surjectivity of π follows from the Lying Over Theorem and the Maximality Theorem 2.2.21, if we can establish that A is integral over AG. Well, for any element f ∈ A, the coeﬃcients of the polynomial

N N−1 Y t + a1t + ··· + aN = (t − g · f) =: Pf (t) g∈G belong to AG, as they are elementary symmetric functions in g · f. On the other hand Pf (f) = 0.

3.6 Products Let X ⊆ An and Y ⊆ Am be algebraic sets. Then the cartesian product n+m X ×Y is an algebraic set in A . Indeed, if we identify k[T1,...,Tm+n] with k[T1,...,Tn]⊗k[T1,...,Tm], then it is easy to see that I(X ×Y ) = I(X) ⊗ k[T1,...,Tm] + k[T1,...,Tn] ⊗ I(Y ) (check it!). From Proposition 2.1.2 we get

k[X × Y ] =∼ k[X] ⊗ k[Y ]. (3.2) 3.7 Rational functions 29 Lemma 3.6.1 Tensor product A ⊗ B of aﬃne k algebras is an aﬃne k-algebra. Moreover, if A and B are domains, then so is A ⊗ B.

Proof The ﬁrst statement follows from (3.2) and Proposition 3.3.3. As- sume A and B are domains and α, α0 ∈ A ⊗ B be such that αα0 = 0. P 0 P 0 0 0 Write α = ai ⊗ bi and α = ai ⊗ bi with the sets {bi} and {bi} each linearly independent. Let M be a maximal ideal in A, anda ¯ de- P P 0 0 note a + M ∈ A/M = k. As ( ai ⊗ bi)( ai ⊗ bi) = 0 in A ⊗ B, in P P 0 0 A/M ⊗ B = k ⊗ B = B we have ( a¯i ⊗ bi)( a¯i ⊗ bi) = 0. As B is 0 domain and the sets {bi} and {bi} are linearly independent, it follows 0 that either all ai ∈ M or all ai ∈ M. Now, recall from Proposition 3.3.3 that A =∼ k[X] for some irreducible variety X. Consider the subvari- 0 0 eties Y and Y of X which are zero sets of the functions {ai} and {ai}, respectively.

Corollary 3.6.2 If X and Y are irreducible then so is X × Y .

Remark 3.6.3 Zariski topology on X × Y is not the product topology of those on X and Y .

Example 3.6.4 This is a generalization of Example 3.5.6(ii). Let X be a closed set in An and f ∈ k[X]. Consider the set X0 ⊆ X × A1 ⊂ An+1 0 ∼ given by the equation Tn+1f(T1,...,Tn) = 1. Note that k[X ] = k[X]f . Then the projection π(T1,...,Tn,Tn+1) = (T1,...,Tn) deﬁned a regular map π : X0 → X. This map deﬁnes a homeomorphism between X0 and the principal open set Xf . This idea will be used to consider a principal open set as an algebraic variety. In fact, it will turn out that k[Xf ] = k[X]f .

3.7 Rational functions In algebraic geometry we need more functions than just globally defined regular functions on a variety X. In fact, if we were planning to deal only with affine algebraic sets such globally defined functions would be ‘enough’ in view of Theorem 3.5.1. However, we will see that constant functions are the only globally defined regular functions on a projective variety. So, as in complex analysis we are going to allow some ‘poles’ and consider functions which are not defined everywhere on X. 30 Affine and Projective Algebraic Sets Definition 3.7.1 Let X be an irreducible algebraic set. The field of fractions of the ring k[X] is denoted k(X) and is called the field of rational functions on X, its elements being rational functions on X.A rational function ϕ ∈ k(X) is regular at the point x ∈ X if it can be f written in the form ϕ = g for f, g ∈ k[X] with g(x) 6= 0. In this case (the f(x) well-defined number) g(x) is called the value of ϕ at x and is denoted ϕ(x).

Note that the set of points on which a rational function ϕ on X is regular is non-empty and open, and hence dense in X. This set is called the domain of ϕ. As the intersection of two non-empty open sets in an irreducible space is non-empty and open again, we can compare a ﬁnite set of rational functions on a non-empty open set. Another useful remark is that a rational function is uniquely determined by its values on a non-empty open set. Indeed, if ϕ = 0 on such a set U, then taking f some presentation ϕ = g for ϕ, we see that f is zero on a non-empty open set U ∩ (X \ Z(g)), which is dense in X, so f = 0.

Theorem 3.7.2 Rational function ϕ regular at all points of an irreducible aﬃne algebraic set X is a regular function on X.

Proof By assumption, for every x ∈ X we can write ϕ(x) = fx(x) for gx(x) fx, gx ∈ k[X] with gx(x) 6= 0. Then the zero set in X of the ideal generated by all functions gx is empty, so by the Nullstellensatz the ideal equals k[X]. So there exist functions h1, . . . hn ∈ k[X] and points Pn x1, . . . , xn ∈ X such that i=1 higxi = 1. Multiplying both sides of f this equality by ϕ (in k(X)) and using the fact that ϕ = xi , we get gx Pn i ϕ = i=1 hifxi , so ϕ ∈ k[X]. The subring of K(X) consisting of all functions regular at the point ∼ x ∈ X is denoted Ox and called the local ring of x. Note that Ox = k[X]Mx , the localization of k[X] at the maximal ideal Mx. So Ox is a local ring in the sense of commutative algebra with the maximal ideal f mx consisting of all rational functions representable in the form g with f(x) = 0 6= g(x). Now Theorem 3.7.2 can be interpreted as

k[X] = ∩x∈X Ox. (3.3)

Informally speaking the local ring Ox describes what happens ‘near the point x’. This becomes a little more clear if we note that Ox is the same as the stalk of rational functions at x: the elements of the stalk are 3.7 Rational functions 31 germs of rational functions at x. One can think of germs as equivalence classes of pairs (U, f), where U is an open set containing x, f is a rational function regular at all points of U, and (U, f) ∼ (V, g) if there is an open set W ⊂ U ∩ V and f|W = g|W . Now, let X ⊆ An be an arbitrary (not necessarily irreducible) algebraic set and U ⊆ X be an open subset. A function f : U → k is regular if for each x ∈ U there exist g, h ∈ k[T1,...,Tn] such that h(x) 6= 0 g and f = h in some open neighborhood of x. The algebra of all regular functions on U is denoted OX (U). Now Ox is defined as the stalk of functions regular in neighborhoods of x. Now, let X ⊆ An be an affine algebraic set and 0 6= f ∈ k[X]. Then the elements of the localization k[X]f can be considered as regular functions on the principal open set Xf (we do imply here that different elements of k[X]f give different functions—check!) We claim that these are precisely all regular functions on Xf :

Theorem 3.7.3 k[X]f is the algebra of regular functions on Xf .

Proof Let g be a regular function on Xf . So we can ﬁnd an open covering of Xf such that on each element U of this covering g equals a b for a, b ∈ k[T1,...,Tn] (with b(x) 6= 0 for all x ∈ U). But principal open sets form a basis of Zariski topology on An, and the topology is ai noetherian. So we may assume that Xf = Xg ∪ · · · ∪ Xg and g = 1 l bi on Xgi for i = 1, . . . , l. Then Xgi ⊆ Xbi . From now on we consider all functions as functions on X via restriction. By the Nullstellensatz, for ni each i, we have gi = bihi for some ni ∈ Z≥0 and hi ∈ k[X]. Note that hi(x) 6= 0 for any x ∈ Xgi , so

ai aihi aihi = = ni . bi bihi gi n n i on Xgi . As Xgi = X i , renaming aihi as ai and g as gi we have that gi i ai g = on Xg . gi i ai aj Now, on Xg ∩ Xg = Xg g we have = , whence aigj − ajgi = 0, i j i j gi gj 2 2 therefore (aigj − ajgi)gigj = 0 everywhere on X. So aigigj = ajgi gj. ai aigi 2 Moreover, on Xgi we have = 2 . Renaming aigi as ai and gi as gi gi ai gi, we are reduced to the case g = on Xg and aigj = ajgi on X. gi i

Now the condition Xf = Xg1 ∪ · · · ∪ Xgl and the Nulstellensatz imply n P f = i cigi for some ci ∈ k[X] for some n. So aj aj X X ajgici X aigjci X gf n|X = f n = c g = = = a c . gj g g i i g g i i j j i i j i j i 32 Aﬃne and Projective Algebraic Sets

n P Since Xgi ’s cover Xf , it follows that gf = i ciai on Xf . So g = P i ciai f n ∈ k[X]f , as required.

3.8 Projective n-space The objects that algebraic geometry can study are much more diverse than just affine algebraic set. To extend our horizons we now demon- strate how projective algebraic sets can be studied. Algebraically, this just means considering homogeneous polynomials instead of all polynomials. Define the projective n-space Pn as the set of equivalence classes on kn+1 \{(0,..., 0)} with respect to the following equivalence relation: (x0, x1, . . . , xn) ∼ (y0, y1, . . . , yn) if and only if there exists an element × c ∈ k such that yi = cxi for all i = 0, 1, . . . , n. Thus every point of Pn has n + 1 coordinates x0, . . . , xn, which are only defined up to a non-zero scalar multiple. To emphasize this fact we will refer to the coordinates of this point as the homogeneous coordinates and denote them by

(x0 : x1 : ··· : xn).

If we want to consider subsets of Pn which are zero sets of polynomials in the homogeneous coordinate functions S0,S1,...,Sn we have to require that these polynomials are homogeneous.

Deﬁnition 3.8.1 Let S be a set of homogeneous polynomials in k[S0,S1,...,Sn]. n A zero of the set S is an element (x0 : x1 : ··· : xn) of P such that f(x0, x1, . . . , xn) = 0 for all f ∈ S. The zero set of S is the set Z(S) of all zeros of S. An algebraic set in Pn (or projective algebraic set) is the zero set of some set of homogeneous polynomials S ⊆ k[S0,S1,...,Sn], in which case S is called a set of equations of the algebraic set.

Note that Z(S) = Z((S)), where (S) is the ideal of k[S0,S1,...,Sn] generated by S. Therefore every algebraic set is the zero set of some homogeneous ideal. Now, by Hilbert’s Basis Theorem, every algebraic set is the sero set of a ﬁnite set of homogeneous polynomials. As in the aﬃne case, one proves that the algebraic sets are closed sets of a topology on Pn, which again is called the Zariski topology. Principal open sets form a base of this topology. The map

n I : {algebraic sets in P } → {homogeneous ideals in k[S0,...,Sn]} 3.8 Projective n-space 33 is deﬁned in the obvious way (you need to check that I(X) is homogeneous!).

Deﬁnition 3.8.2 The ideal M0 of k[S0,...,Sn] generated by S0,...,Sn is called the superﬂuous ideal. The following projective version of the Nullsellensatz follows easily from the classical one.

Theorem 3.8.3 (Projective Nullstellensatz) The maps I and Z induce an order-reversing bijection between algebraic sets in Pn and non-superﬂuous homogeneous radical ideals in k[S0,...,Sn]. Under this correspondence, irreducible algebraic sets correspond to the prime ideals.

n Let Ui ⊂ P be the subset consisting of all points with non-zero ith homogeneous coordinate. This is the principal open set corresponding n to the function Si. We call the Ui (the ith) aﬃne open set in P . The terminology is justiﬁed by the following. The map

αi :(x0, . . . , xn) 7→ (x0/xi, . . . , xi−1/xi, xi+1/xi, . . . , xn/xi)

n is a bijection between Ui and A . We will refer to the functions

Tj :(x0, . . . , xn) 7→ xj/xi, (j = 0, . . . , i − 1, i + 1, . . . , n) as the aﬃne coordinates on Ui. We claim that αi is not just a bijection but a homeomorphism between n Ui and A . Indeed, to each polynomial f(T0,...,Ti−1,Ti+1,...,Tn) we associate its homogenization ˆ deg f f(S0,...,Sn) := Si f(S0/Si,...,Si−1/Si,Si+1/Si,...,Sn/Si), which is clearly a homogeneous polynomial in S0,...,Sn. Now, if X in n A is the zero set of polynomials f1, . . . , fm ∈ k[T0,...,Ti−1,Ti+1,...,Tn], then −1 ˆ ˆ α (X) = Ui ∩ Z(f1,..., fm). ˆ ˆ n −1 We note in passing, that Z(f1,..., fm) is the closure in P of α (X) (why?). Conversely, to each homogeneous polynomial g(S0,...,Sn) we associate the polynomial

g¯(T0,...,Ti−1,Ti+1,...,Tn) := g(T0,...,Ti−1, 1,Ti+1,...,Tn). Now

α(Z(g1, . . . , gl) ∩ Ui) = Z(¯g1,..., g¯l). 34 Affine and Projective Algebraic Sets Lemma 3.8.4 (Affine Criterion) Let X be a topological space with an open cover X = ∪i∈I Ui, and Y ⊆ X. Then Y is closed if and only n Y ∩ Ui is closed in Ui for all i. In particular, a subset Y of P is closed if and only if its intersection Y ∩ Ui with the ith affine open set is closed in Ui for all i.

Proof The ‘only-if’ part is obvious. For the ‘if’ part, by assumption n each Y ∩ Ui = Zi ∩ Ui for some closed set Zi in P . It suﬃces to check that n Y = ∩i∈I (Zi ∪ (P \ Ui)).

Well, let y ∈ Y and i ∈ I. Either y ∈ Ui and then y ∈ Y ∩ Ui ⊂ Zi, or n n n y ∈ P \Ui. Conversely, if y ∈ Zi ∪(P \Ui) for all i. As P = ∪Ui, there n is an i with y ∈ Ui. Then y 6∈ P \Ui, hence y ∈ Zi, and x ∈ Zi ∩Ui ⊂ Y .

3.9 Functions A rational expression f = p(S0,...,Sn) can be considered as a function on q(S0,...,Sn) n P (defined at the points where q(S0,...,Sn) 6= 0) only if p and q are homogeneous of the same degree, in which case we will refer to f as a rational function of degree 0. Let X ⊂ Pn be a projective algebraic set, p x = (x0, . . . , xn) ∈ X, and f = q be of degree 0. If q(x0, . . . , xn) 6= 0, then we say that f is regular at x. If a degree 0 rational function is regular at x, then it is also regular on some neighborhood of x. For any set Y ⊆ Pn, a function f on Y is called regular if for any x ∈ Y there exists a rational function g regular at x and such that f = g on some open neighborhood of x in Y . If U is an open subset of X we write OX (U) for the set of all regular functions on U. We will prove later that the only functions regular on projective algebraic sets are constants. This underscores the importance of considering rational functions regular only on some open subsets. Let U be an open subset of Pn contained in some affine open set n Ui. Then U is also open in Ui, which is canonically identified with A . n n We claim that OP (U) = OA (U). Indeed, assume for example that n i = 0, and let f ∈ OP (U). This means that there is an open cover U = n pj (S0,...,Sn) W1 ∪· · ·∪Wl in and rational functions defined on Wj such P qj (S0,...,Sn) pj pj (1,T1,...,Tn) that f = on Wj, j = 1, . . . , l. Then we also have f = on qj qj (1,T1,...,Tn) Wj, where T1,...,Tn are the affine coordinates on U0. Conversely, let

n f ∈ OA (U). This means that there is an open cover U = V1 ∪ · · · ∪ Vm 3.10 Product of projective algebraic sets 35

gj (T1,...,Tn) gj of U and rational functions defined on Vj such that f = hj (T1,...,Tn) hj deg h S j gˆ on V , j = 1, . . . , m. Now, on V we can also write f = 0 j , where j j deg gj ˆ S0 hj ˆ gˆj and hj are homogenizations. n Let X ⊆ P be a projective algebraic set, and U0,...,Un be the affine n open sets in P . Put Vi := X∩Ui. Then X = V0∪· · ·∪Vl is an open cover of X. Moreover, Vi an affine algebraic set in Ui, and Ui is canonically identified with An. Let U be an open subset of X which is contained in some Vi. Then U is an open subset of Vi. The argument as in the previous paragraph can be modified to prove the following more general result: a function on U is regular in the sense of the projective algebraic set X if and only if it is regular in the sense of the affine algebraic set

Vi, i.e. OX (U) = OVi (U).

3.10 Product of projective algebraic sets Let X ⊂ Pn and Y ⊂ Pm be projective algebraic sets. We would like to consider X × Y as a projective algebraic set in a natural way. For example, we could have X = Pn and Y = Pm. It is quite clear that there is no natural identiﬁcation of Pn ×Pm with Pn+m (play with that!). But there is a natural Segre embedding of Pn × Pm into P(n+1)(m+1)−1: n m (n+1)(m+1)−1 ϕ : P × P → P ,

((T0,...,Tn), (S0,...,Sm)) 7→ (T0S0,...,T0Sm,...,TnS0,...,TnSm) It is easy to see that ϕ is injective. We next show that im ϕ is closed (n+1)(m+1)−1 in P . Let wij, 0 ≤ i ≤ n, 0 ≤ j ≤ m be the homogeneous coordinates in P(n+1)(m+1)−1. We claim that im ϕ is the zero set of the following equations:

wijwkl = wkjwil (0 ≤ i, k ≤ n, 0 ≤ j, l ≤ m). (3.4) That all points of im ϕ satisfy these equations is clear. Conversely, if the numbers wij satisfy these equations, and wkl 6= 0, then

(··· : wij : ... ) = ϕ(x, y) where x = (w0l : ··· : wnl) and y = (wk0 : ··· : wkm). So, we have proved that the image of Pn ×Pm under the Segre embedding is a projective algebraic set, and this is what we will understand by the product of Pn and Pm. More generally, let X be an algebraic set in Pn and Y be an algebraic set in Pm. By the product of X and Y we understand ϕ(X × Y ), which we show to be algebraic. Well, if X is given 36 Aﬃne and Projective Algebraic Sets by the equations Fα(T0,...,Tn) = 0 and Y is given by the equations Gβ(S0,...,Sm) = 0, then X × Y is the zero set of the equations (3.5) together with Fα(w0j, . . . , wnj) for 1 ≤ j ≤ m and Gβ(wi0, . . . , wim) for 1 ≤ i ≤ n.

3.11 Example: Grassmann varieties and ﬂag varieties Let V be an n-dimensional vector space. As a set, the Grassmann variety Gr(V ) (or Gr(n)) is just the set of all r-dimensional (linear) subspaces in V . However, we need to explain how is Gr(V ) a projective algebraic set. Of course, we already know that for r = 1 when Gr(V ) is nothing but the projective space P(V ) = Pn−1. In general we are going to realize r Gr(V ) as an algebraic set in the projective space P(Λ (V )). Deﬁne the map

r ψ : Gr(V ) → P(Λ (V )) as follows. Let l1, . . . , lr be a basis of a subspace L ⊂ V . Then ψ(L) is r defined to be the span of the vector l1 ∧ · · · ∧ lr ∈ Λ (V ). It is easy to check that ψ is a well defined embedding. We claim that the image of ψ is an algebraic set. In order to see that, let us fix a basis {v1, . . . , vn} of V . Then the basis of Λr(V ) is

{vi1 ∧ · · · ∧ vir | 1 ≤ i1 < ··· < ir ≤ n}.

Denote the vi1 ∧ · · · ∧ vir -coeﬃcient of l1 ∧ · · · ∧ lr by µi1...ir . Then the homogeneous coordinates of ψ(L) are (··· : µi1...ir : ... ). These homogeneous coordinates are called the Pl¨ukker coordinates of L. We accept the following convention: given a collection of numbers {µi1...ir |

1 ≤ i1 < ··· < ir ≤ n} we assume that µi1...ir are also deﬁned for any i1, . . . , ir with 1 ≤ i1, . . . , ir ≤ n in such a way that after two indices are interchanged, µi1...ir gets multiplied by −1; in particular, if two indices are the same, it is zero. With these assumptions the Pl¨ukker coordinates can be described as Pn follows. Write li = j=1 aijvj. Then µi1...ir is the determinant of the matrix formed by the columns of A := (aij) with indices i1, . . . , ir.

Theorem 3.11.1 Numbers µi1...ir are Plükker coordinates of some r- dimensional subspace L ⊂ V if and only if they are not simultaneously zero and if for all i1, . . . , ir+1, j1, . . . , jr−1 the following relation (called 3.11 Example: Grassmann varieties and flag varieties 37 Plükker relation) holds:

r+1 X k (−1) µ µi j ...j = 0. i1...ibk...ir+1 k 1 r−1 k=1

Proof Expanding the determinant µikj1...jr−1 along the ﬁrst column, we obtain r X µikj1...jr−1 = asik Ns, s=1 where Ns does not depend on k. Thus, it suﬃces to prove that

r+1 X k (−1) µ asi = 0 (3.5) i1...ibk...ir+1 k k=1 for all s. Add the sth row to A to obtain an (r + 1) × n matrix As. Then the left hand side of (3.5) is, up to a sign, the expansion of the determinant of the matrix formed by the columns of As with indices i1, . . . , ir+1 along the last row. But this determinant is zero.

Conversely, assume that µi1...ir are not simultaneously zero and the Pl¨ukker relations hold. It suﬃces to prove that there exists an r × n matrix A such that

µi1...ir = Mi1...ir (1 ≤ i1, . . . , ir ≤ n), (3.6)

where Mi1...ir is the minor formed by the columns of A with indices i1, . . . , ir. We may assume that µ1...r = 1. We will look for A in the form   1 0 ... 0 a1,r+1 . . . a1n 0 1 ... 0 a2,r+1 . . . a2n   . ......  ......  0 0 ... 1 ar,r+1 . . . arn

r−i Note that for j > r we have M1...î...rj = (−1) aij. Thus , we must r−i set aij = (−1) µ1...î...rj, in which case the equality (3.6) holds at least for the sets {i1, . . . , ir} which differ from {1, . . . , r} in no more than one element.

Now it remains to prove that (3.6) holds if the set {i1, . . . , ir} differs from {1, . . . , r} in m elements for any m. We use induction on m. We may assume that i1 6∈ {1, . . . , r}. Then, using the Plükker relations, we 38 Affine and Projective Algebraic Sets get

r X µ = µ µ = (−1)k+1µ µ . (3.7) i1...ir 1...r i1...ir i11...k...rˆ ki2...ir k=1

On the other hand, it follows from the ﬁrst part of the theorem that the same condition holds for the minors of A:

r X M = (−1)k+1M M . (3.8) i1...ir i11...k...rˆ ki2...ir k=1

By the induction hypothesis, the right hand sides of (3.7) and (3.8) coincide. Therefore Mi1...ir = µi1...ir .

A ﬂag in the n-dimensional vector space V is a chain

0 ⊂ V1 ⊂ V2 ⊂ · · · ⊂ Vn = V of subspaces with dim Vi = i for all i = 1, . . . , n. Let F(V ) be the set of all ﬂags in V . This set can be given a natural structure of a projective algebraic set called ﬂag variety. Note that Vi ∈ Gi(V ), so we can consider F(V ) as a subset of G1(V ) × · · · × Gn(V ), and we claim that this is a closed subset.

Indeed, it suffices to prove that the condition for Vd to be contained in Vd+1 is a closed condition for each d. In checking that we may forget about other spaces and work in P(Λd(V )) × P(Λd+1(V )). Let us apply Affine Criterion. The open covering we are going to use is the direct products of the affine open sets in P(Λd(V )) and P(Λd+1(V )). The d affine open sets in P(Λ (V )) are given by conditions µi1...id 6= 0. As they are all the same we may work with the set U given by µ1...r 6= 0. Then Vd ∈ U if and only if Vd is spanned by the vectors of the form Pn ∼ d(n−d) vi + j=d+1 aijvj, i = 1, . . . , d. In fact, U ∩ Gd(V ) = A and the aij can be considered as the affine coordinates on U ∩ Gd(V ). Now, 0 d+1 let U be the affine open set in P(Λ (V )) containing Vd+1 given by

µi1...id+1 6= 0. As Vd ⊂ Vd+1, we must have that i1 = 1, . . . , id = d, for otherwise the intersection with U × U 0 is empty. We may also assume without loss of generality that id+1 = d + 1. Now, Vd+1 is spanned by Pn the vectors of the form vi + j=d+2 bijvj, i = 1, . . . , d+1. In fact, the bij 0 can be considered as the aﬃne coordinates on U ∩ Gd+1(V ). Now the condition that Vd is contained in Vd+1 can be written by the polynomial equations aij = bij + ai,d+1bd+1,i for all 1 ≤ i ≤ d and d + 2 ≤ j ≤ n. 3.12 Example: Veronese variety 39 3.12 Example: Veronese variety

Consider all homogeneous polynomials of degree m in S0,S1,...,Sn. n+m They form a vector space of dimension m . The corresponding pro- νn,m n+m νn,m jective space is P where νn,m := m − 1. To each point of P there corresponds a hypersurface of degree m in Pn (since proportional polynomials deﬁne the same hypersurface). νn,m Denote the homogeneous coordinates in P by vi0...in for all tuples (i0, . . . , in) of non-negative integers with i0 + ··· + in = m. Consider the n νn,m map αm : P → P , deﬁned by

i0 in vi0...in (αm((a0 : ··· : an))) = a0 . . . an . (3.9)

The map is well-deﬁned, as among the monomials in the right hand side m of (3.9) there are ai which all turn into 0 only if all ai = 0. The map n αm is clearly injective. It is called Veronese map, and αm(P ) is called Veronese variety. Formulas (3.9) imply that all points of the Veronese variety satisfy equations

vi ...i vj ...j = vk ...k vl ...l 0 n 0 n 0 n 0 n (3.10) if i0 + j0 = k0 + l0, . . . , in + jn = kn + ln.

Conversely, it follows from the relations (3.10) that at least one of the coordinates of the form v0...m...0 is non-zero. Indeed, assume otherwise, and prove by induction on the amount k of non-zeros among {i0, . . . , in} that all vi0...in = 0. The induction base k = 1 follows from our assumption. On the other hand, let k ≥ 2 and assume that the statement is true for k − 1. Let ir be the minimal non-zero element in {i0, . . . , in} and is be the minimal non-zero element of {i0, . . . , in}\{ir}. Now, the relation

v2 = v v i0...ir ...is...in i0...0...is+ir ...in i0...ir ...is−ir ...in is among the relations (3.9). By the inductive assumption, the right hand side of it is zero, so vi0...in is also zero, completing the induction step.

Now, let, for example, vm0...0 6= 0. Then our point with homogeneous coordinates (vi0...in ) is the image under the Veronese map of the point with coordinates

u0 = vm0...0, u1 = vm−1,1,0...0, . . . , un = vm−1,0...0,1. 40 Aﬃne and Projective Algebraic Sets Indeed, it suﬃces to check that

i0 i1 in (vm0...0) (vm−1,1,0...0) ... (vm−1,0...0,1) m−1 = vi0...in . vm0...0 or, equivalently,

i0−m+1 i1 in (vm0...0) (vm−1,1,0...0) ... (vm−1,0...0,1) = vi0...in . (3.11) We prove this by induction on the lexicographical order on the tuples (i0 . . . in). For the highest tuple (m0 ... 0) the result is obvious. Every other (i0 . . . in) has some ir 6= 0. Now,

vi0...ir ...in vm0...0 = vi0+1...ir −1...in vm−1...1...0. (3.12)

If vm−1...1...0 = 0, it follows that vi0...in = 0, in which case (3.11) is clear. Otherwise, substituting (3.12) into (3.11), we reduce (3.11) for (i0, . . . in) to (3.11) for (i0 + 1, . . . , ir − 1, . . . , in), which is true by induction. P i0 in Let F = ai0...in u0 . . . un be a form of degree m and H be a hy- n persurface in P deﬁned by the equation F = 0. Then αm(H) is the n intersection of αm(P ) with the hyperplane given by the equation X ai0...in vi0...in = 0. Let us now concentrate on the special case

1 3 3 2 2 3 α3 : P → P :(a0 : a1) 7→ (a0 : a0a1 : a0a1 : a1). The corresponding Veronese variety C is called the twisted cubic. It is described by the equations

F0 = F1 = F2 = 0, (3.13) where

2 2 F0 = v30v12 − v21,F1 = v21v12 − v30v03,F2 = v21v03 − v12. The twisted cubic consists of all points of the form (1 : c : c2 : c3) for c ∈ k together with the point (0 : 0 : 0 : 1). Let Qi be the hypersurfaces described by Fi = 0. Then C = Q0 ∩ Q1 ∩ Q2, but C 6= Qi ∩ Qj for any two hypersurfaces Qi and Qj. In fact the following beautiful geometric fact is true: the intersection Qi ∩Qj equals C ∪Lij for some (projective) line Lij (it is easy to see that no line is contained in C). In order to prove this we consider a more general problem. For λ = (λ0 : λ1 : λ2) ∈ P2 deﬁne the hypersurface Qλ by Fλ, where

Fλ := λ0F0 + λ1F1 + λ2F2. 3.13 Problems 41

We claim that for λ 6= µ, one has Qλ ∩Qµ = C ∪Lλ,µ for some line Lλ,µ. Note that the equations (3.13) are equivalent to the requirement that the matrix v30 v21 v12 v21 v12 v03 has rank less than 2. Now note that Fλ is the determinant of the matrix   v30 v21 v12 v21 v12 v03 . λ2 λ1 λ0

So the locus outside of C of Fλ = Fµ = 0 is the rank ≤ 2 locus of the matrix   v30 v21 v12 v21 v12 v03   ,  λ2 λ1 λ0  µ2 µ1 µ0 which as λ and µ are linearly independent is the same as the locus of

v30 v21 v12 v21 v12 v03

λ2 λ1 λ0 = λ2 λ1 λ0 = 0,

µ2 µ1 µ0 µ2 µ1 µ0 which is a line.

3.13 Problems

Problem 3.13.1 True or false? Let I,J be ideals in k[T1,...,Tn]. Then Z(I) ∪ Z(J) = Z(IJ). √ √Solution. True. By the Nullstellensatz,√ it suﬃces√ to prove that I ∩ J = √IJ. Well, IJ ⊂ I ∩ J implies IJ ⊂ I ∩ J. Conversely, let x ∈ I ∩ J. Then xn ∈ I ∩ J, whence x2n ∈ IJ.

Problem√ √ 3.13.2 True or false? Let I,J be ideals in k[T1,...,Tn]. Then I ∩ J = IJ.

Solution. True. See the previous problem.

Problem 3.13.3 Let I and J be ideals of A = C[x, y] and Z(I)∩Z(J) = ∅. Show that A/(I ∩ J) =∼ A/I × A/J. 42 Aﬃne and Projective Algebraic Sets Solution. In view of the Chinese Remainder Theorem, we need only to show that I + J = A. Otherwise, let M be a maximal ideal containing 2 I + J. By the Nullstellensatz, M = Ma for some a ∈ C . Then a ∈ Z(I) ∩ Z(J).

Problem 3.13.4 True or false? Any decreasing sequence of algebraic sets in An stabilizes. Solution. True by the Nullstellensatz and Hilbert Basis Theorem.

Problem 3.13.5 True or false? Any increasing sequence of algebraic sets in An stabilizes. Solution. False. Take ”increasing sets of points”.

Problem 3.13.6 If X = ∪Uα is an open covering of an algebraic set, then X = Uα1 ∪ · · · ∪ Uαl for some α1, . . . , αl. Solution. Otherwise we would have an inﬁnite strictly decreasing sequence of closed subsets, which contradicts Problem 3.13.4.

Problem 3.13.7 True or False?

(i) {(x, y) ∈ A2 | x2 + y2 = 1} is homeomorphic to k (in Zariski topology). (ii) The set k \{(0)} with induced Zariski topology is not homeomorphic to any variety.

Solution. (i) True. Our variety has the same cardinality as k and coﬁnite topology, see Lemma 2.1.1 (even characteristic 2 is O.K., because then Z(x2 + y2 − 1) = Z(x + y − 1)). (ii) False. This set and k have the same cardinality and coﬁnite topology.

Problem 3.13.8 True or false? A system of polynomial equations

f1(T1,...,Tn) = 0 . .

fm(T1,...,Tn) = 0 over k has no solutions in An if and only if 1 can be expressed as a linear P combination 1 = i pifi with polynomial coeﬃcients pi. 3.13 Problems 43

Solution. True. The ﬁrst condition is equivalent to (f1, . . . , fm) = k[T ], in view of the Nullstellensatz.

Problem 3.13.9 Let char k 6= 2. Decompose Z(x2 + y2 + z2, x2 − y2 − z2 + 1) into irreducible components. Solution. An easy calculation shows that Z(x2 +y2 +z2, x2 −y2 −z2 +1) equals √ √ Z(x = i/ 2, y2 + z2 = 1/2) ∪ Z(x = −i/ 2, y2 + z2 = 1/2), union of two irreducible sets, since y2 + z2 = 1/2 is an irreducible polynomial.

Problem 3.13.10 True or false? The Zariski topology on Am+n is the product topology of the Zariski topologies on Am and An. Solution. False. Consider the case m = n = 1.

Problem 3.13.11 Let k have characteristic p > 0, and Fr : k → k, a 7→ ap be the Frobenius homomorphism. True or false: (i) Fr is a homeomorphism in the Zariski topology. (ii) Fr is an isomorphism of algebraic sets. Solution. (i) is true, as Fr is a bijection. (ii) is false as Fr∗ is not an isomorphism.

Problem 3.13.12 Prove that the hyperbola xy = 1 and k are not isomorphic. Solution. If ψ : k[x, y]/(xy − 1) → k[T ] is an isomorphism, then ψ(x) and ψ(y) must be invertible, which leads to a contradiction.

Problem 3.13.13 For the regular map f : A2 → A2, (x, y) 7→ (x, xy) describe im f. Is the image dense in A2? Open? Closed? Solution. The image is A2 \{(0, b) | b 6= 0}. It is dense because it contains a non-empty open set x 6= 0. So it is not closed. It is also not open, as the origin belongs to the closure of the complement C (in fact, I(C) = (x)).

Problem 3.13.14 Let X consist of two points. Prove that k[X] =∼ k⊕k. 44 Aﬃne and Projective Algebraic Sets Solution. Use the Nullstellensatz and the Chinese Remainder Theorem (cf. Problem 3.13.3).

Problem 3.13.15 Describe all automorphisms of the algebraic set k. Solution. All automorphisms are linear of the form x 7→ ax + b with a 6= 0. This follows by considering automorphisms of k[T ]. By the way, the automorphism group is isomorphic to the semidirect product of k× and k.

Problem 3.13.16 The graph of a morphism ϕ : X → Y of aﬃne algebraic sets is a closed set in X × Y isomorphic to X.

Solution. Let s1, . . . , sn be coordinate functions on Y . Then the graph ∗ is the zero locus of the functions ϕ (si) ⊗ 1 − 1 ⊗ si ∈ k[X] ⊗ k[Y ] = k[X × Y ]. Next, check that the maps x 7→ (x, f(x)) and (x, f(x)) 7→ x are morphisms between X and the graph which are inverse to each other.

Problem 3.13.17 Let ϕ : X → Y be a morphism of aﬃne algebraic sets. Show that inverse image of a principal open set in Y is a principal open set in X.

−1 Solution. ϕ (Yf ) = Xϕ∗(f).

Problem 3.13.18 Let X,X0 be topological spaces. (i) A subspace Y ⊆ X is irreducible if and only if Y¯ is irreducible. (ii) If ϕ : X → X0 is a continuous map and X is irreducible, then ϕ(X) is irreducible. Solution. See Humphreys.

Problem 3.13.19 Let ϕ : X → Y be a regular map. Then ϕ(X) is dense in Y if and only if ϕ∗ is injective. Give an example when ϕ(X) is dense in Y but ϕ(X) 6= Y . Solution. I(im ϕ) = {g ∈ k[Y ] | g(ϕ(x)) = 0 for any x ∈ X} = {g ∈ k[Y ] | ϕ∗(g) = 0} = ker ϕ∗. Now the result follows from Z(I(im ϕ)) = im ϕ. For the example see Problem 3.13.13.

Problem 3.13.20 Let X,Y ⊂ Ar be closed subsets, and ∆ ⊂ A2r be the diagonal, i.e. a subset given by equations T1 = S1,...,Tr = Sr. If z ∈ X ∩ Y define ϕ(z) = (z, z). Prove that ϕ defines an isomorphism from X ∩ Y onto (X × Y ) ∩ ∆. 3.13 Problems 45 Solution.(x, y) 7→ x defines the inverse morphism.

Problem 3.13.21 True or false? Let X be an aﬃne algebraic set with irreducible components X1,...,Xl. Then a function f on X is in k[X] if and only if f|Xi ∈ k[Xi] for all i.

Solution. This is actually false! Let X = X1 ∪ X2, where X1 is the line 2 2 2 in A given by x = 0, and X2 ⊂ A is the parabola x = y . Consider 2 the function f which is 0 on X1, and which maps the point (y , y) of X2 to y. Then clearly f|X1 and f|X2 are regular. Now assume that there is a polynomial F (x, y) with F |X = f. Since F |X1=0, it follows 2 that F (x, y) = xg1(x, y) + x g2(x, y) + ... . Now, F |X2 = f|X2 gives 2 2 2 4 2 y = F (y , y) = y g1(y , y) + y g2(y , y) + ... , which is impossible by degrees. 4 Varieties

4.1 Affine varieties In this section we will define affine varieties which can be thought of as a ‘coordinate-free version’ of affine algebraic sets and functions on them.

Deﬁnition 4.1.1 A sheaf of functions on a topological space X is a function F which assigns to every non-empty open subset U ⊂ X a k-algebra F(U) of k-valued functions on U (with respect to the usual point-wise operations) such that the following two conditions hold: (i) If U ⊂ V are two non-empty open sets and f ∈ F(V ), then the restriction f|U ∈ F(U). (ii) Given a family of open sets Ui, i ∈ I, covering U and functions fi ∈ F(Ui) for each i ∈ I, such that fi and fj agree on Ui ∩ Uj, there must exist a function f ∈ F(U) whose restriction to Ui equals fi.

Deﬁnition 4.1.2 A topological space X together with a sheaf of functions OX is called a geometric space. We refer to OX as the structure sheaf of the geometric space.

Deﬁnition 4.1.3 Let (X, OX ) and (Y, OY ) be geometric spaces. A morphism

f :(X, OX ) → (Y, OY ) is a continuous map f : X → Y such that for every open subset U of Y and every ϕ ∈ OY (U) the function f ∗(ϕ) := ϕ ◦ f

−1 belongs to OX (f (U)).

46 4.1 Aﬃne varieties 47 Remark 4.1.4 We will often use a shorthand f : X → Y for the morphism f :(X, OX ) → (Y, OY ).

Example 4.1.5 Let X be an aﬃne or a projective algebraic set. To each non-empty open subset U ⊂ X we assign the ring OX (U) which consists of all regular functions on U. Then (X, OX ) is a geometric space. Moreover the notion of a morphism agrees with the one we had before (think about it!).

Let (X, OX ) be a geometric space and Z be a subset of X with induced topology. We can make Z into a geometric space by defining OZ (V ) for an open V ⊂ Z as follows: f : V → k is in OZ (V ) if and only if there exists an open covering V = ∪iVi in Z such that for each i we have f|Vi = gi|Vi for some gi ∈ OX (Ui) where Ui is an open subset of X containing Vi. It is not difficult to see that OZ is a sheaf of functions on Z (see it!). We will refer to it as the induced structure sheaf and denote it by OX |Z. Note that if Z is open in X then a subset V ⊂ Z is open in Z if and only if it is open in X, and OX (V ) = OZ (V ). Let X be a topological space and X = ∪iUi be its open cover. Given sheaves of functions OUi on Ui for each i, which agree on each Ui ∩ Uj, we can define a natural sheaf of functions OX on X by ‘gluing’ the OUi . Let U be an open subset in X. Then OX (U) consists of all functions on

U, whose restriction to each U ∩ Ui belongs to OUi (U ∩ Ui). If x ∈ X we denote by evx the map from functions on X to k obtained by evaluation at x:

evx(f) = f(x).

Deﬁnition 4.1.6 A geometric space (X, OX ) is called an aﬃne (algebraic) variety if the following three conditions hold:

(i) k[X] := OX (X) is a ﬁnitely generated k-algebra, and the map

X → Homk−alg(k[X], k), x 7→ evx is a bijection. (ii) For each 0 6= f ∈ k[X] the set

Xf := {x ∈ X | f(x) 6= 0} is open, and every non-empty open set in X is a union of some Xf . (iii) OX (Xf ) = k[X]f . 48 Varieties Example 4.1.7 It follows from the results of chapter 3 (in particular, Theorem 3.7.3) that affine algebraic sets with sheaves of regular functions are affine varieties. We claim that, conversely, every affine variety is isomorphic (as a geometric space) to an affine algebraic set with the sheaf of regular functions. Indeed, let (X, OX ) be an affine variety. Since k[X] is a finitely generated algebra of functions, we can write

k[X] = k[T1,...,Tn]/I for some radical ideal I. By the property (i) of affine varieties and the Nulstellensatz, we can identify X with Z(I) as a set, and k[X] with the regular functions on Z(I). The Zariski topology on Z(I) has the principal open sets as its base, so it now follows from (ii) that the identification of X and Z(I) is a homeomorphism. Finally, by (iii), OX (Xf ) and the regular functions on the principal open set Xf are also identified. This is enough to identify OX (U) with regular functions on U for any open set U, as regularity is a local condition.

Remark 4.1.8 The argument of Example 4.1.7 shows that the affine variety can be recovered completely from its algebra A := k[X] of regular functions, and conversely. We make it precise as follows. Define a functor F from the category of affine varieties to the category of affine ∗ algebras via F(X) = k[X] := OX (X), F(f) = f . We now describe a quasi-inverse functor G from the affine algebras to the affine varieties (this means that F ◦ G =∼ Id and G ◦ F =∼ Id, i.e. F and G establish an equivalence of categories, see Problem 4.6.1. In particular, if (X, OX ), (Y, OY ) are affine varieties and f : X → Y is a map, then f is a morphism if and only if f ∗ maps k[Y ] to k[X], and f : X → Y is an isomorphism if and only if f ∗ is an isomorphism from k[Y ] to k[X]. So let A be an affine k-algebra. We define G(A) to be the affine variety Specm A = (X, OX ), where

X := Homk-alg(A, k)

(which in view of Hilbert’s Nulstellensatz, can be identiﬁed with the set of the maximal ideals of A, whence the name). Note that the elements of a can be considered as k-valued functions on X via

f(x) := x(f)(f ∈ A, x ∈ X = Homk-alg(A, k)).

Now consider the topology on X whose basis consists of all Xf := {x ∈ X | f(x) 6= 0} for f ∈ A. In order to deﬁne a structure sheaf on the 4.2 Prevarieties 49 topological space X, set

OX (Xf ) := Af (f ∈ A \{0})

(again elements of Af can be considered as functions on Xf in a natural way). Now for any U = ∪f Xf a function on U is in OX (U) if and only if its restriction to each Xf is in OX (Xf ).

Example 4.1.9 In view of Example 4.1.7, a closed subset of an aﬃne variety is an aﬃne variety (as usual, with the induced sheaf), cf. Prob- lem 4.6.6.

Example 4.1.10 If (X, OX ) is an affine variety, then it is easy to check that the principal open set Xf is also an affine variety (think why this does not contradict what was claimed in Example 4.1.7.) On the other hand, not every open subset of X is an affine variety, see Problem 4.6.4.

4.2 Prevarieties

Deﬁnition 4.2.1 An (algebraic) prevariety is a geometric space (X, OX ) such that X has an open covering X = U1 ∪ · · · ∪ Ul, and each geometric space (Ui, OUi ) with the induced structure sheaf OUi is an aﬃne variety.

Example 4.2.2 In view of §3.9, each projective algebraic set with the sheaf of regular functions is a prevariety. We will refer to varieties isomorphic to projective algebraic sets with sheaves of regular functions as projective varieties.

Lemma 4.2.3 Let (X, OX ) be a prevariety with aﬃne open covering X = U1 ∪ · · · ∪ Ul. (i) X is a noetherian topological space. (ii) Any open subset U of X is again a prevariety. (iii) Any closed subset Z of X is again a prevariety.

Proof (i) follows from the fact that each Ui is noetherian. (ii) As U = ∪i(U ∩ Ui), it suffices to prove that each U ∩ Ui has an affine open covering. But U ∩ Ui is an open subset of an affine Ui, so it is a union of the principal open sets in Ui, which are affine by Example 4.1.10. 50 Varieties

(iii) Z = ∪i(Z ∩ Ui), and closed subsets Z ∩ Ui of aﬃne varieties are aﬃne. A subset of a topological space is called locally closed if it is an intersection of an open set and a closed set. It follows from above that a locally closed subset of a prevariety is again a prevariety. We will refer to the locally closed subsets as subprevarieties.

Theorem 4.2.4 (Aﬃne Criterion) Let X,Y be prevarieties, and ϕ : X → Y be a map. Assume that there is an aﬃne open covering Y = ∪i∈I Vi and an open covering X = ∪i∈I Ui such that

(i) ϕ(Ui) ⊂ Vi for each i ∈ I; (ii) f ◦ ϕ ∈ OX (Ui) whenever f ∈ OY (Vi). Then ϕ is a morphism.

Proof An affine open covering of X induces that of each Ui. So, by extending the index set if necessary we reduce to the case where Ui are also affine. Now by assumption, ϕi := ϕ|Ui : Ui → Vi is a morphism of affine varieties. In particular, ϕi is continuous, whence ϕ is continuous. −1 Let V ⊂ Y be an open subset, f ∈ OY (V ), and U := ϕ (V ). By (ii), −1 −1 f ◦ϕ ∈ OX (ϕ (V ∩Vi)). But ϕ (V ∩Vi) ⊇ U ∩Ui, so f ◦ϕ ∈ OX (U ∩Ui) for all i. Now, since U is the union of the U ∩Ui and since OX is a sheaf, f ◦ ϕ ∈ OX (U). Let X be an irreducible prevariety. Consider pairs (U, f) where U is an open subset of X and f ∈ OX (U). We call two such pairs (U, f) and (U 0, f 0) equivalent if there is a non-empty open subset V ⊂ U ∩ U 0 such that f|V = f 0|V (in which case we will also have f|(U ∩ U 0) = f 0|(U ∩ U 0)). It is easy to check using the irreducibility of X that this defines an equivalence relation. Moreover, the set of equivalence classes is a field with respect to the obvious operations. (For example, (U, f)−1 = (U ∩ Uf , 1/f)). This field is called the field of rational functions on X and denoted k(X). It is easy to see that if X is affine then this definition agrees with the one we had before. Moreover, if U ⊂ X is a non-empty open subset, then k(X) = k(U). Let F be a sheaf of functions on a topological space X and x ∈ X. The open sets in X containing x form inverse system with respect to inclusion. The stalk Fx of F at x is defined to be the corresponding limit of algebras

Fx = lim F(U). U 4.3 Products 51

The elements of the stalk Fx are called germs of functions at x. One can think of germs as equivalence classes of pairs (U, f), where U is an open set containing x, f ∈ F(U), and (U, f) ∼ (V, g) if there is an open set W ⊂ U ∩ V containing x such that f|W = g|W . If (X, OX ) is a prevariety, we write simply Ox for (OX )x and call it the local ring of x. It is easy to see that the ring Ox is local in the sense of commutative algebra. Its unique maximal ideal is denoted mx—it consists of the germs of functions equal to zero at x. If X is an irreducible affine variety, this definition agrees with the one given in §3.7. Note also that Ox is a ‘local notion’, which means that if x ∈ U for an open subset U ⊂ X, and OU is the induced sheaf on U, then Ox defined using U is the same as the one defined using X.

4.3 Products

Theorem 4.3.1 Finite products exist in the category of prevarieties.

Proof It suﬃces to deal with two prevarieties (X, OX ) and (Y, OY ). We need to prove that there exists a prevariety (Z, OZ ) together with morphisms π1 : Z → X and π2 : Z → Y such that the following universal property holds: if (W, OW ) is another prevariety with morphisms ϕ1 : W → X and ϕ2 : W → Y , then there exists a unique morphism ψ : W → Z such that πiψ = ϕi for i = 1, 2. For any set S denote by Map(S, k) the algebra of all functions from S to k. Observe that for any open U ⊂ X and V ⊂ Y the natural map of algebras

OX (U) ⊗ OY (V ) → Map(U × V, k). is injective. So we will identify elements of OX (U)⊗OY (V ) as functions on U × V . Now deﬁne a topology on the set X × Y by saying that the open sets will be the unions of the sets of the form

(U × V )h := {x ∈ U × V | h(x) 6= 0}, where U ⊂ X, V ⊂ Y are arbitrary open subsets and h ∈ OX (U) ⊗ OY (V ). We will refer to such (U × V )h as principal open sets. Checking that this is a topology boils down to

0 0 0 0 (U × V )h ∩ (U × V )h0 = ((U ∩ U ) × (V ∩ V ))hh0 . 52 Varieties Next we deﬁne a structure sheaf on X × Y . Let W be an open set in X × Y and f ∈ Map(W, k). Then we say that f is regular if and only if there is an open cover of W by the principal open sets (U × V )h so that on each of them we have a f|(U × V ) = h hm for some a ∈ OX (U) ⊗ OY (V ) and some non-negative integer m. 0 This deﬁnes a sheaf OX×Y . Indeed, let W ⊂ W be an open subset. 0 0 0 We have W = ∪(U × V )h and W = ∪(U × V )h0 . So

0 0 0 0 0 W = ∪((U × V )h ∩ (U × V )h0 ) = ∪((U ∩ U ) × (V ∩ V ))hh0 .

Moreover,

0m a 0 0 (a↓)h |((U ∩ U ) × (V ∩ V )) 0 = , hm hh (hh0)m where a↓ denotes the restriction of a from U × V to (U ∩ U 0) × (V ∩ V 0), 0 0 0 which belongs to OX (U ∩ U ) ⊗ OY (V ∩ V ). So f|W is regular. The second axiom of sheaf is obvious.

Now we want to show that (X × Y, OX×Y ) is a prevariety. First, it is easy to see that for the case where X,Y are affine, our definition agrees with the one from §3.6. So if X = ∪iUi, Y = ∪jVj are open affine covers, then X × Y = ∪i,jUi × Vj is an open affine cover.

Let π1 : X × Y → X and π2 : X × Y → Y be the natural projections, and let us check the universal property. First of all, we need to check that the projections are morphisms. They are continuous: for example, for an open U ⊂ X, we have π−1(U) = U × V , which is open. Moreover, ∗ ∗ let f ∈ OX (U). Then (π1 (f))(x, y) = f(x). So π1 (f) = f ⊗ 1 ∈ OX (U) ⊗ OY (Y ) is regular.

Finally, let ϕ1 : W → X and ϕ2 : W → Y be morphisms. It is clear that if ψ required in the universal property exists, then it must send w ∈ W to (ϕ1(w), ϕ2(w)). To show that ψ is a morphism, we use the affine criterion. We know that the products U × V of the affine open 0 −1 −1 subsets cover X × Y . Open subsets of the form W = ϕ1 (U) ∩ ϕ2 (V ) ∗ P 0 cover W , and ψ maps a function ai ⊗ ai from OX×Y (U × W ) to P ∗ ∗ 0 0 the function ϕ1(ai)ϕ2(ai) ∈ OW (W ). By the affine criterion, ψ is a morphism. 4.4 Varieties 53 4.4 Varieties

Definition 4.4.1 A prevariety X is called an (algebraic) variety if the diagonal ∆(X) = {(x, x) | x ∈ X} is closed in X × X. An equivalent condition is as follows: for any prevariety Y and any two morphisms ϕ, ψ : Y → X the set {y ∈ Y | ϕ(y) = ψ(y)} is closed in Y . Indeed, applying this condition to π1, π2 : X × X → X we conclude that ∆ is closed; conversely, the preimage of ∆ under ϕ×ψ : Y → X ×X is {y ∈ Y | ϕ(y) = ψ(y)}. It follows from the previous paragraph that a subprevariety of a variety is variety. We will refer to it a subvariety from now on. In the category of topological spaces with usual product topology on X × X the Definition 4.4.1 is equivalent to the Hausdorff axiom. So we can think of varieties as prevarieties with some sort of an unusual Hausdorff axiom.

Example 4.4.2 An example of a prevariety which is not a variety is given by the aﬃne line with a doubled point, see Problem 4.6.7.

Lemma 4.4.3 Let Y be a variety and X be a prevariety. (i) If ϕ : X → Y is a morphism, then the graph

Γϕ := {(x, ϕ(x)) | x ∈ X} is closed in X × Y . (ii) If ϕ, ψ : X → Y are morphisms which agree on a dense subset of X then ϕ = ψ.

Proof (i) Γϕ is the inverse image of ∆(Y ) with respect to the morphism X × Y → Y × Y, (x, y) → (ϕ(x), y). (ii) The set of all points where ϕ and ψ agree is closed.

Lemma 4.4.4 Aﬃne varieties are varieties.

Proof Note that

∆(X) = {(x, y) ∈ X × X | evx = evy} = {(x, y) ∈ X × X | f(x) = f(y) for all f ∈ k[X]} = Z(f ⊗ 1 − 1 ⊗ f | f ∈ k[X]}. 54 Varieties Lemma 4.4.5 The product of two varieties is a variety.

Proof Under the isomorphism (X × Y ) × (X × Y )→ ˜ (X × X) × (Y × Y ), ∆(X × Y ) maps to ∆(X) × ∆(Y ), which is closed.

Lemma 4.4.6 Let X be a prevariety. If every pair of points x, y ∈ X lie in an open aﬃne subset, then X is a variety.

Proof Let Y be a prevariety and ϕ, ψ : Y → X be morphisms. Set Z := {y ∈ Y | ϕ(y) = ψ(y)}. In oprder to show that Z is closed, let z ∈ Z¯, and x1 = ϕ(z), x2 = ψ(z). By assumption, x1 and x2 lie in an open aﬃne subset V of X. Then U := ϕ−1(V ) ∩ ψ−1(V ) is an open neighborhood of z, which must have a non-trivial intersection with Z. But Z ∩ U = {y ∈ U | ϕ0(y) = ψ0(y)} where ϕ0, ψ0 : U → V are restrictions of ϕ, ψ to U. As V is a variety, Z ∩ U is closed in U. So U \ (Z ∩ U) is open subset whose intersection with Z is empty. Hence z ∈ Z.

It follows easily from Lemma 4.4.6 that projective varieties are varieties, see Problem 4.6.16.

4.5 Dimension Recall that we have assigned to every irreducible variety its field of rational functions k(X). As k(X) is a finitely generated field extension of k, it has a finite transcendence degree tr. degk k(X) over k. This degree is called the dimension of X and denoted dim X. In general dimension of X is defined as the maximum of the dimensions of its irreducible components.

Example 4.5.1 (i) dim An = dim Pn = n. (ii) Dimension of a finite set is 0. Conversely, if dim X = 0, then X is finite. Indeed, let X be an irreducible affine variety X ⊂ An n of dimension 0. Let t1, . . . , tn be coordinates on A considered as functions on X. Then ti are algebraic over k, so can take only finitely many values. So X is finite.

Example 4.5.2 Grassmann variety Gr(n) is covered by the open subsets r(n−r) µi1...ir 6= 0, isomorphic to A , so dim Gr(n) = r(n − r). 4.5 Dimension 55 Proposition 4.5.3 Let X and Y be irreducible varieties of dimensions m and n, respectively. Then dim X × Y = m + n.

Proof We may assume that X and Y are affine. Let s1, . . . , sp and t1, . . . , tq be generators of the algebras k[X] and k[Y ], respectively. Then s1, . . . , sp and t1, . . . , tq generate the fields k(X) and k(Y ) over k, respectively. So we can choose transcendence bases out of them. After renum- bering, if necessary, transcendence bases are s1, . . . , sm and t1, . . . , tn. Recall that k[X ×Y ] = k[X]⊗k[Y ]. Let us write si for si ⊗1 and tj for 1 ⊗ tj. As s1, . . . , sp, t1, . . . , tq generate k[X × Y ], they also generate the field k(X × Y ) over k. Moreover, these generators depend algebraically on s1, . . . , sm, t1, . . . , tn. So it suffices to prove that s1, . . . , sm, t1, . . . , tn are algebraically independent.

Assume there is an algebraic dependence f(s1, . . . , sm, t1, . . . , tn) = 0. Then for each ﬁxed x ∈ X the function f(s1(x), . . . , sm(x), t1, . . . , tn) is zero on Y . As t1, . . . , tn are algebraically independent, all coeﬃcients g(s1(x), . . . , sm(x)) of the polynomial f(s1(x), . . . , sm(x),T1,...,Tn) ∈ k[T1,...,Tn] are zero. As x was arbitrary and s1, . . . , sm are algebraically independent, it follows that the polynomial g(S1,...,Sm) ∈ k[S1,...,Sm] is zero. Hence f(S1,...,Sm,T1,...,Tn) = 0.

Proposition 4.5.4 Let X be an irreducible variety and Y be a proper closed subvariety. Then dim Y < dim X.

Proof We may assume that Y is irreducible and that X is aﬃne, say of dimension d. Let A = k[X], A¯ = k[Y ]. Then A¯ = A/P for some non- zero prime ideal P of A. The transcendence bases of k(X) and k(Y ) can be found in A and A¯. Assume that dim Y ≥ d. Then we can choose d algebraically independent elementsa ¯1,..., a¯d ∈ A¯. These elements are cosets of some a1, . . . , ad ∈ A which are of course also algebraically independent. Let b ∈ P be a non-zero element. As dim X = d, there must exist a non-trivial algebraic dependence f(b, a1, . . . , ad) = 0, where f(T0,T1,...,Td) ∈ k[T0,T1,...,Td]. Since b 6= 0 we may assume that T0 does not appear in all monomials of the polynomial f, i.e. the polynomial g(T1,...,Tn) = f(0,T1,...,Tn) is non-zero. But then g(¯a1,..., a¯d) = 0, giving a contradiction.

Corollary 4.5.5 Let X be an irreducible aﬃne variety and Y is an irreducible closed subvariety of codimension 1. Then Y is a component of the variety Z(f) for some f ∈ k[X]. 56 Varieties Proof By assumption Y 6= X, so there exists a non-zero function f ∈ k[X] with f|Y = 0. Then Y ⊆ Z(f) ( X. Let Z be an irreducible component of Z(f) containing Y . By Proposition 4.5.4, dim Z < dim X. So dim Z = dim Y , and by Proposition 4.5.4 again, Y = Z.

Lemma 4.5.6 If X is an irreducible aﬃne variety for which k[X] is a u.f.d., then every closed subvariety of codimension 1 has form Z(f) for some f ∈ k[X].

Proof Let Y be the subvariety, and Y1,...,Yl be the components of Y . Then I(Y ) = ∩I(Yi). So, if we can prove that I(Yi) = (fi), then I(Y ) = (f1 . . . fl) (as the fi must be powers of diﬀerent irreducible elements). Thus we may suppose that Y is irreducible. Let P = I(Y ), a non-zero prime ideal in k[X]. It therefore contains an irreducible element f. So (f) is a prime ideal contained in P . If (f) ( P , then Y = Z(P ) ( Z (f) ( X, which contradicts the assumption that codimension of Y is 1, thanks to Proposition 4.5.4.

Remark 4.5.7 The statement of Lemma 4.5.6 fails if k[X] is not a 4 u.f.d. For example, let X = Z(T1T4 − T2T3) ⊂ A . It contains the 0 planes L and L given by the equations T2 = T4 = 0 and T1 = T3 = 0, respectively. Clearly, L ∩ L0 = {(0, 0, 0, 0)}. We claim that L is not Z(f) for any f ∈ k[X]. Otherwise, Z(f|L0) = {(0, 0, 0, 0)}, which is impossible, because it has codimension 2 in Z0.

If X is an aﬃne variety and f ∈ k[X] is a non-invertible element, then the zero set Z(f) is called a hypersurface in X. If k[X] is a u.f.d., the irreducible components of this hypersurface are precisely hypersurfaces deﬁned by the irreducible components of f.

Proposition 4.5.8 All irreducible components of a hypersurface in An have codimension 1.

Proof It suﬃces to consider the zero set X of an irreducible polynomial p(T1,...,Tn). We may assume that (say) Tn appears in p, as p is non-scalar. Let ti := Ti|X. So k(X) = k(t1, . . . , tn). In view of Proposition 4.5.4 it suﬃces to prove that t1, . . . , tn−1 are algebraically independent.

Assume that there is a non-trivial polynomial relation g(t1, . . . , tn−1) = 4.5 Dimension 57

0, so the polynomial g(T1,...,Tn−1) is zero on X. It follows that g is divisible by p, which is impossible since Tn appears in p. The proof of the following more general fact requires more powerful commutative algebra:

Theorem 4.5.9 Let X be an irreducible aﬃne variety, 0 6= f ∈ k[X] be a non-invertible elemet, and Y be an irreducible component of Z(f). Then Y has codimension 1 in X.

Proof Let Y1,...,Yl be the components of Z(f) diﬀerent from Y , and P := I(Y ),Pi := I(Yi) be the corresponding (prime) ideals in k[X]. As the intersection of prime ideals is radical, it follows from the Nullstel- lensatz that p (f) = P ∩ P1 ∩ · · · ∩ Pl.

Note by the Nullstellensatz that P 6⊃ P1 ∩· · ·∩Pl. Take g ∈ P1 ∩· · ·∩Pl with g 6∈ P . Note that Xg is an irreducible affine variety of the same dimension as X, and, by the choice of g, Y ∩ Xg is the zero set of f in Xg. On the other hand, Y ∩ Xg is a principal open sunset of Y , so it suffices to prove that its codimension in Xg is 1. So from the very beginning we may assume that Y = Z(f) and P = p(f). Now, apply Noether’s Normalization Lemma 2.2.28 to the domain R := k[X]: R is integral over some subring S isomorphic to k[T1,...,Td], where d = dim X. Let E = k(X) and F be the field of fractions of S. Then E/F is finite (generated by fnitely many algebraic elements). By

Corollary 2.2.27, the norm map NE/F takes values in S on elements of R.

Denote NE/F (f) =: f0 ∈ S. We claim that f0 ∈ P . Let irr (f, F ) = k k−1 x + a1x + ··· + ak ∈ S[x], see Lemma 2.2.26. By Lemma 2.2.25, f0 m is ±ak for some m. Now f0 ∈ (f) ⊆ P , in view of k k−1 m−1 0 = (f + a1f + ··· + ak)ak k−1 m−1 k−2 m−1 m−1 = f(f ak + a1f ak + ··· + ak−1ak ) ± f0.

Let Q be the radical of the ideal (f0) in S. Then Q ⊆ S ∩ P . We claim that Q = S ∩ P . Indeed, let g ∈ S ∩ P . Since g ∈ P , we have gl = fh for some l ∈ N and h ∈ R. Computing the norms, we get l[E:F ] g = NE/F (f)NE/F (h) = f0NE/F (h).

As NE/F (h) ∈ S, we deduce that g ∈ Q. 58 Varieties We conclude that Q is a prime ideal in S. Since S is a UFD, it follows that f0 is a power of an irreducible polynomial p in S, whence Q = (p). Clearly p is not a scalar. Considering S as the algebra of regular functions on Ad, we now conclude that Z(Q) is an irreducible hypersurface of codimension 1, thanks to Proposition 4.5.8. So the transcendence degree of the quotient field of S/Q over k is d − 1. On the other hand, R is integral over S implies that R/P is integral over S/(P ∩ S) = S/Q. So the quotient field of R/P also has transcendence degree d − 1 over k. But the last quotient field is k(Y ), so dim Y = d − 1.

Corollary 4.5.10 Let X be an irreducible variety, U be an open subset of X, and f ∈ OX (U) be a non-invertible element. Then every irreducible component of the zero set of f in U has codimension 1 in X.

Proof Let Y be an irreducible component of the zero set of f in U, and V be an aﬃne open subset in X contained in U with Y ∩V 6= ∅. Then using Theorem 4.5.9, we have dim Y = dim(Y ∩ V ) = dim V − 1 = dim X − 1.

Corollary 4.5.11 Let X be an irreducible variety, and Y ⊆ X be an irreducible closed subset of codimension r. Then there exist irreducible closed subsets Yi of codimension 1 ≤ i ≤ r, such that Y = Yr ⊂ Yr−1 ⊂ · · · ⊂ Y1.

Proof By passing to the aﬃne open subset which intersects Y , we may assume that X is aﬃne. Apply induction on r. If r = 1, there is nothing to prove. Since Y 6= X, there exists a function f 6= 0 in I(Y ), and Y lies in an irreducible component Y1 of Z(f). By Theorem 4.5.9, codim Y1 = 1, and we can apply induction.

Corollary 4.5.12 (Topological Characterization of Dimension) The dimension of an irreducible variety X is the largest integer d for which there exist a chain of non-empty irreducible closed subsets

X0 ( X1 ( ··· ( Xd = X.

Proof This follows from Corollary 4.5.11 and the fact that the dimension of a proper closed subset of a variety is strictly smaller than the dimension of the variety. 4.5 Dimension 59 Remark 4.5.13 The topological characterization shows that, when X is irreducible aﬃne, dim X is the Krull dimension dim k[X] of k[X], i.e. the maximal length d of the chain of prime ideals 0 ( P0 ( P1 ( ··· ( Pd ( k[X]. Now Theorem 4.5.9 can be restated as follows: let A be an aﬃne k-algebra which is a domain, and f ∈ A be neither zero nor a unit, and let P be a prime ideal minimal among those containing (f); then dim A/P = dim A−1. This statement is a version Krull’s principal ideal theorem.

Corollary 4.5.14 Let X be an irreducible variety, f1, . . . , fr ∈ OX (X). Then each irreducible component of the set Z(f1, . . . , fr) has codimension at most r.

Proof Apply Theorem 4.5.9.

n Remark 4.5.15 Let X = A , f1 = T1, f2 = T1 +1. Then Z(f1, f2) = ∅, which is of codimension ∞, because by agreement dim ∅ = −∞. Think why this does not contradict Corollary 4.5.14.

Corollary 4.5.16 Let X be an irreducible aﬃne variety, and Y ⊂ X be a closed irreducible subset of codimension r ≥ 1. Then Y is a component of Z(f1, . . . , fr) for some f1, . . . , fr ∈ k[X].

Proof We prove more generally that for closed irreducible subsets Y1 ⊃ Y2 ⊃ · · · ⊃ Yr with codim Yi = i there exist functions fi ∈ k[X] such that all components of Z(f1, . . . , fi) have codimension i, and Yi is one of those components (1 ≤ i ≤ r). This is indeed a more general statement in view of Corollary 4.5.11. Apply induction on i. For i = 1 we use Corollary 4.5.5 to ﬁnd a function f1 such that Y1 is a component of Z(f1), and then Theorem 4.5.9 to deduce that all components of Z(f1) have codimension 1. Assume that the functions f1, . . . fi−1 have been found, and let Yi−1 = Z1,Z2,...,Zm be the irreducible components of Z(f1, . . . , fi−1). Each of them has codimension i − 1, so none of them lies in Yi. So I(Zj) 6⊃ I(Yi) for all j = 1, . . . , m. The ideals I(Zj) are prime, so it follows from Theorem 2.1.5 that their union also does not contain in I(Yi). Let fi be a function which is zero on Yi but which is not identically zero on all Zj. If Z is a component of Z(f1, . . . , fi), then Z lies in one of the components Zj of the set Z(f1, . . . , fi−1), and also in Z(fi). So Z is a component of Z(fi) ∩ Zj, which by Theorem 4.5.9, has codimension 1 in 60 Varieties

Zj, and hence codimension i in X. Finally, the function fi is zero on Yi, and Yi has codimension i, so Yi is one of the components of Z(f1, . . . , fi).

Remark 4.5.17 The statement shows that for a prime ideal P in an aﬃne k-algebra which is a domain, if P has height r , then there exist elements f1, . . . , fr such that P is minimal among the prime ideals containing (f1, . . . , fr).

Remark 4.5.18 A closed subvariety X of An (resp. Pn) of codimension r is called a set theoretic complete intersection if there exist r polynomials fi ∈ k[T1,...,Tn] (resp. r homogeneous polynomials fi ∈ k[S0,S1,...,Sn]) such that X = Z(f1, . . . , fr). Moreover, X is called an ideal theoretic complete intersection if the fi can be chosen so that I(X) = (f1, . . . , fr).

4.6 Problems

Problem 4.6.1 Prove that the functors F :(X, OX ) 7→ k[X] and G : A 7→ Specm A are quasi-inverse equivalences of categories between aﬃne varieties over k and aﬃne k-algebras (this means FG =∼ Id and GF =∼ Id).

Solution. To prove that FG =∼ Id, let A be an affine k-algebra. By definition, k[Specm A] =∼ A, where A is considered as an algebra of functions on Specm A via a(x) = x(a), see Remark 4.1.8. It is easy to see that the isomorphism k[Specm A] =∼ A is natural. Now, let (X, OX ) be an affine variety. It follows from the axioms of the affine variety and the definition of Specm k[X] that

X → Specm k[X], x 7→ evx is an isomorphism of varieties, which is clearly natural. So GF =∼ Id.

Problem 4.6.2 True or false? Let X be a prevariety and U ⊂ X is a non-empty open subset. If f ∈ OX (U) then f is a morphism from the prevariety U to k = A1.

Problem 4.6.3 Principal open sets in aﬃne varieties are aﬃne varieties.

Problem 4.6.4 Prove that A2 \{(0, 0)} is not an affine variety. 4.6 Problems 61 Problem 4.6.5 Intersection of affine open subsets is affine.

Problem 4.6.6 Prove that a closed subset of an affine variety is again an affine variety without using affine algebraic sets.

Problem 4.6.7 Make sense out of Example 4.4.2.

Problem 4.6.8 The product of irreducible prevarieties varieties is irreducible.

Problem 4.6.9 Let ϕ1 : X1 → Y1 and ϕ2 : X2 → Y2 be morphisms of prevarieties. Then ϕ1 × ϕ2 : X1 × X2 → Y1 × Y2, (x1, x2) 7→ (ϕ1(x1), ϕ2(x2)) is also a morphism of prevarieties.

Problem 4.6.10 Let X,Y be prevarieties. Prove that the projections X × Y → X and X × Y → Y are open maps, i.e. map open maps to open maps. Do they have to map closed sets to closed sets?

Problem 4.6.11 Let ϕ : X → Y be prevarieties. Prove that the projection π1 induces an isomorphism from Γϕ ⊂ X × Y onto X.

Problem 4.6.12 Let X,Y be prevarieties, and X0 ⊂ X,Y 0 ⊂ Y be subprevarieties. Explain how X0 × Y 0 can be considered as a subprevariety of X × Y .

Problem 4.6.13 Prove that any morphism P1 → A1 must be constant.

Problem 4.6.14 Let f : A1 → A1 be a morphism. Then there is a unique extension morphism f˜ : P1 → P1 such that f|A1 = f.

Problem 4.6.15 Show that every isomorphism f : P1 → P1 is of the ax+b form f(x) = cx+d for some a, b, c, d ∈ k, where x is the coordinate on A1.

Problem 4.6.16 Prove that Pn is a variety.

Problem 4.6.17 Prove that the Veronese embedding is an isomorphism of Pn onto its image.

Problem 4.6.18 Let X ⊂ Pn be a projective algebraic set considered as a variety and f ∈ k[S0,...,Sn] be a non-constant homogeneous poly- 62 Varieties nomial. Then X \ Z(f) is an aﬃne variety. (Hint: Reduce to the case where f is linear using the Veronese embedding).

Problem 4.6.19 Prove that the product of projective varieties deﬁned in §3.10 using Segre embedding is the categorical product.

Problem 4.6.20 Irreducible closed subvarieties of a variety X satisfy A.C.C.

Problem 4.6.21 The dimension of a linear subvariety of An (that is a subvariety deﬁned by linear equations) has the value predicted by linear algebra.

Problem 4.6.22 Let X and Y be closed subvarieties of An. For any non- empty irreducible component Z of X∩Y , we have codim Z ≤ codim X+ codim Y .

Problem 4.6.23 Fill in the details for Example 4.5.2

Problem 4.6.24 Prove that X × {point} =∼ X. 5 Morphisms

5.1 Fibers A ﬁber of a morphism ϕ : X → Y is a subset of the form ϕ−1(y) for y ∈ Y . As ϕ is continuous, ﬁbers of ϕ are closed subvarieties in Y . Of course ϕ−1(y) is empty if y 6∈ im ϕ. If X is irreducible and ϕ(X) is dense in Y we say that the morphism ϕ is dominant. In this case Y will also have to be irreducible, as the image of an irreducible topological space under a continuous map is irreducible and the closure of an irreducible subspace is irreducible. More generally, if X is not necessarily irreducible, then a morphism ϕ : X → Y is dominant, if ϕ maps every component of X onto a dense subset of some component of Y , and im ϕ is dense in Y . If ϕ is a dominant morphism of irreducible varieties then the comorphism ϕ∗ induces an embedding of k(Y ) into k(X). In particular, dim X ≥ dim Y . Let ϕ : X → Y be a morphism, and W ⊆ Y be an irreducible closed subset. If the restriction of ϕ to an irreducible component Z of ϕ−1(W ) is dominant as a morphism from Z to W , then we say that Z dominates W . If im ϕ ∩ W is dense in W then at least one of the components of ϕ−1(W ) dominates W .

Theorem 5.1.1 Let ϕ : X → Y be a dominant morphism of irreducible varieties, and let r = dim X − dim Y . Let W be a closed irreducible subset of Y , and Z be a component of ϕ−1(W ) which dominates W . Then dim Z ≥ dim W +r. In particular, if y ∈ im ϕ, then the dimension of each component of the ﬁber ϕ−1(y) is at least r.

Proof Let U be an aﬃne open subset of Y which intersects W . Then

63 64 Morphisms U ∩ W is dense in W , and hence irreducible. Also, ϕ(Z) ∩ U is dense in W . So for the purpose of comparing dimensions we can consider U instead of Y , W ∩U instead of W , ϕ−1(U)∩Z instead of Z, and ϕ−1(U) instead of X. Thus we may assume that Y is aﬃne.

Let s = codim Y W . By Corollary 4.5.16, W is an irreducible compo- ∗ nent of Z(f1, . . . , fs) for some f1, . . . , fs ∈ k[Y ]. Setting gi = ϕ (fi) ∈ OX (X), we have Z ⊆ Z(g1, . . . , gs). As Z is irreducible, it actually lies in some component Z0 of Z(g1, . . . , gs). But by assumption W = ϕ(Z), and ϕ(Z) ⊆ ϕ(Z0) ⊆ Z(f1, . . . , fs). As W is a component of Z(f1, . . . , fs), it −1 follows that ϕ(Z) = ϕ(Z0) = W , whence Z0 ⊆ ϕ (W ). But Z is a com- −1 ponent of ϕ (W ), so Z = Z0, i.e. Z is a component of Z(g1, . . . , gs). In view of Corollary 4.5.14, codim X Z ≤ s. The theorem follows. The theorem says that the non-empty ﬁbers of a morphism are not ‘too small’. The following example shows that they can be ‘too large’.

Example 5.1.2 Let ϕ : A2 → A2 be the morphism given by ϕ(x, y) = (x, xy). Then ϕ is dominant. The ﬁber ϕ−1((0, 0)) is the y-axis, so it is 1-dimensional. On the other hand, all other non-empty ﬁbers have the ‘right’ dimension 0.

5.2 Finite morphisms Let ϕ : X → Y be a morphism of affine varieties. If the ring k[X] is integral over the subring ϕ∗(k[Y ]), then we say that the morphism ϕ is finite. The main case is when X and Y are irreducible and ϕ is dominant and finite. Then we can consider k[Y ] as a subring of k[X] and then k(Y ) as a subfield of k(X). Moreover, since k[X] is integral and finitely generated over k[Y ], k(X) is a finite algebraic extension of k(Y ), so dim X = dim Y . Fibers of finite maps are finite sets (which explains the terminology). n Indeed, let ϕ : X → Y be finite, X ⊂ A , and t1, . . . , tn be the coor- n dinates on A as functions on X. By definition, each ti satisfies some k ∗ k−1 ∗ equation of the form ti + ϕ (a1)ti + ··· + ϕ (ak) = 0 with ai ∈ k[Y ]. Let y ∈ Y and x ∈ ϕ−1(y). Then

k k−1 ti(x) + a1(y)ti(x) + ··· + ak(y) = 0, which has only finitely many roots. Note that the morphism ϕ from Example 5.1.2 is dominant but not ∗ finite. Indeed, T2 is not integral over ϕ (k[Y ]) = k[T1,T1T2]. 5.2 Finite morphisms 65 Remark 5.2.1 If ϕ : X → Y is a surjective morphism of irreducible affine varieties, and all fibers are finite, then it can be proved that ϕ is finite, see [Sp, §5.2]. We will not pursue this now.

Example 5.2.2 Let X be an affine variety and G be a finite group of automorphisms of X, whose order N is prime to char k. We claim that the projection map π : X → X/G is finite, cf. Example 3.5.7.

Theorem 5.2.3 Let ϕ : X → Y be a ﬁnite morphism of aﬃne varieties with f(X) dense in Y . Then ϕ(X) = Y .

Proof Let y ∈ Y , and let My be the corresponding maximal ideal of k[Y ]. If t1, . . . , tn are the coordinate functions on Y and y = (a1, . . . , an), then −1 My = (t1 − a1, . . . , tn − an). Deﬁning equations of the variety ϕ (y) ∗ ∗ −1 are ϕ (t1) = a1, . . . , ϕ (tn) = an, and ϕ (y) is empty if and only if ∗ ∗ (ϕ (t1) − a1, . . . , ϕ (tn) − an) = k[X]. If we identify k[Y ] with the subring of k[X] via ϕ∗, the last condition is equivalent to the condition Myk[X] = k[X]. Note that k[X] is a ﬁnitely generated k[Y ]-module in view of Propo- sition 2.2.7(ii). So by Corollary 2.1.7, Myk[X] 6= k[X].

Corollary 5.2.4 Finite maps are closed, i.e. they map closed sets onto closed sets. To be more precise, let ϕ : X → Y be a ﬁnite map, and let Z ⊂ X be a closed subset. Then ϕ|Z : Z → ϕ(Z) is ﬁnite. In particular, ϕ(Z) = ϕ(Z).

Proof We may assume that ϕ(X) = Y . Denote R = k[X],S = k[Y ]. As ϕ∗ is injective, we can identify S with a subring of R, and then R is integral over S, since ϕ is finite. If I is an ideal of R then R/I ⊃ S/(I∩S) is another integral ring extension. Let I = I(Z). Then ϕ(Z) = Z0, where Z0 = Z(I ∩ S). Moreover, I0 := I ∩ S is radical, so I0 = I(Z0). The affine algebras of Z and Z0 are R/I and R/I0, so the remarks in the previous paragraph show that ϕ|Z : Z → Z0 is again finite and dominant. It remains to apply Theorem 5.2.3 to this map.

Corollary 5.2.5 Let ϕ : X → Y be a ﬁnite dominant morphism of irreducible aﬃne varieties. Suppose that k[Y ] is integrally closed. If W is a closed irreducible subset of Y and Z is any component of ϕ−1(W ), then ϕ(Z) = W . 66 Morphisms Proof Keep the notation of the proof of Corollary 5.2.4, and let J = I(W ). Then I ∩ S = I(ϕ(Z)) = I(ϕ(Z)) and I is a minimal prime ideal of R for which I∩S ⊇ J. It follows from the Going Down Theorem 2.2.29 that I ∩ S = J. So ϕ(Z) = W .

5.3 Image of a morphism Let S ⊂ R be two finitely generated domains over k with quotient fields 0 E ⊂ F . Set r := tr. degE F . Let R be the localization of R with respect to the multiplicative system S∗ of non-zero elements of S. Note that F is also field of fractions of R0. On the other hand R0 contains E, so it can be considered as an E-algebra. By Noether’s normalization lemma, R0 is integral over a ring E[T1,...,Tr] for some algebraically independent elements T1,...,Tr over E. Note that T1,...,Tr can be chosen in R, as all possible denominators are in E. 0 Now compare the integral extension E[T1,...,Tr] ⊂ R with the extension S[T1,...,Tr] ⊂ R. The latter extension is not necessarily integral, but R is finitely generated over S as a ring. Moreover, each gener- ator of R satisfies a monic polynomial equation over E[T1,...,Tr]. If f is a common denominator of all coefficients appearing in such equations for all generators, then it is clear that Rf is integral over Sf [T1,...,Tr] (and T1,...,Tr are algebraically independent over Sf , because they are algebraically independent even over E). These remarks will be used in the proof of the following theorem.

Theorem 5.3.1 Let ϕ : X → Y be a dominant morphism of irreducible varieties, and r = dim X − dim Y . Then (i) im ϕ contains an open subset U of Y . (ii) if all local rings of points of Y are integrally closed, then we can choose U in part (i) so that it has the following property: if W ⊂ Y is an irreducible closed subset which meets U, and Z is a component of ϕ−1(W ), which meets ϕ−1(U), then dim Z = dim W + r.

Proof By passing to an open affine subset of Y , we may assume that Y is affine (cf. the proof of Theorem 5.1.1). We may also reduce to the case where X is affine. Indeed, let X = ∪Vi be an open affine covering. As Vi is dense in X, we have ϕ(Vi) is dense in X, so the restriction ϕ|Vi : Vi → Y is a dominant morphism of irreducible affine varieties. 5.3 Image of a morphism 67

Now, if Ui is an open subset of Y as in (i) or (ii) for ϕ|Vi, then U = ∩iVi satisfies (i) and (ii), respectively. Let R = k[X],S = k[Y ]. Consider S as a subring of R in a usual way, and find elements T1,...,Tr ∈ R, f ∈ S, such that Rf is integral over Sf [T1,...,Tr]. Recall that Rf = k[Xf ] and Sf = k[Yf ]. So the ∼ affine algebra Sf [T1,...,Tr] = Sf ⊗ k[T1,...,Tr] can be considered as r k[Yf × A ]. Then the restriction ϕ|Xf : Xf → Yf can be decomposed ψ r π1 as a composition Xf −→ Yf × A −→ Yf where ψ is a finite dominant −1 morphism. Set U = Yf and note that ϕ (U) = Xf . Moreover, ψ is surjective by Theorem 5.2.3, and π1 is obviously surjective, so U ⊆ ϕ(X), which proves (i).

To prove (ii), we also set X = Xf , U = Y = Yf . Then as above ϕ = π1 ◦ ψ, where ψ is a ﬁnite morphism. It follows from the assumption and (3.3) that the ring k[Y ] = Sf is integrally closed. Now by Theorem 2.2.15, Sf [T1,...,Tr] is also integrally closed. If W is a closed irreducible subset of Y and Z is any component of ϕ−1(W ), then Z is a component of ψ−1(W × Ar). Hence ψ(Z) = W × Ar, and dim Z = dim ψ(Z) = dim W + r, see Corollary 5.2.5.

In (ii) above it would be enough to assume that local rings are integrally closed only for some non-empty open subset of Y (we could pass from Y to this open subset in the very beginning of the proof). It will later turn out that this condition is always satisﬁed, see Theorem 6.3.1. So the assumption can actually be dropped.

Proposition 5.3.2 Let ϕ : X → Y be a bijective morphism of irreducible varieties. Then dim X = dim Y , and there are open subsets U ⊂ X and V ⊂ Y such that ϕ(U) = V and ϕ|U : U → V is a ﬁnite morphism.

Proof We may assume that Y is affine. Let W ⊂ X be an open affine subset. As W is dense in X, we have ϕ(W ) is dense in Y , so the restriction ϕ|W : W → Y is a dominant morphism of irreducible affine varieties. Let R = k[W ],S = k[Y ]. Consider S as a subring of R ∗ via (ϕ|W ) , and find elements x1, . . . , xr ∈ R, f ∈ S, such that Rf is integral over Sf [x1, . . . , xr]. Recall that Rf = k[Wf ] and Sf = k[Yf ]. So ∼ the affine algebra Sf [x1, . . . , xr] = Sf ⊗ k[x1, . . . , xr] can be considered r as k[Yf ×A ]. Then the restriction ϕ|Wf : Wf → Yf can be decomposed ψ r π1 as a composition Wf −→ Yf × A −→ Yf where ψ is a finite dominant morphism. Now, ψ is surjective by Theorem 5.2.3. Hence ϕ|Wf : Wf → 68 Morphisms

Yf is surjective, and hence bijective by our assumption. This is only possible if r = 0, so ϕ|Wf : Wf → Yf is ﬁnite, and dim X = dim Y .

Recall that a subset of a topological space is called locally closed if it is an intersection of an open set and a closed set. A ﬁnite union of locally closed sets is called a constructible set.

Theorem 5.3.3 Let ϕ : X → Y be a morphism of varieties. Then ϕ maps constructible sets onto constructible sets. In particular, im ϕ is constructible.

Proof Locally closed subset of a variety is itself a variety, so it suﬃces to prove that im ϕ is constructible. We can also assume that X and Y are irreducible. Apply induction on dim Y . If dim Y = 0, there is nothing to prove. By inductive assumption, we may assume that ϕ is dominant. Let U be an open subset contained in im ϕ, see Theorem 5.3.1(i). Then the irreducible components Wi of Y \ U have dimensions less than −1 dim Y . By induction, the restriction of ϕ to Zi := ϕ (Wi) has image constructible in Wi, so also constructible in Y . Now, ϕ(X) is a union of U and the constructible sets ϕ(Zi), so ϕ(X) is also constructible.

Proposition 5.3.4 Let ϕ : X → Y be a dominant morphism of irreducible varieties.

(i) The set {y ∈ Y | dim ϕ−1(y) ≥ n} is closed for any n.

(ii) For x ∈ X let εϕ(x) denote the maximal dimension of any component of the set ϕ−1(ϕ(x)) containing x. Then for all n ≥ 0, the set En(ϕ) := {x ∈ X | εϕ(x) ≥ n} is closed in X.

Proof We prove (ii), the proof of (i) is very similar (and easier). Ap- ply induction on dim Y , the case dim Y = 0 being clear. Let r = dim X − dim Y , and let U be an open subset contained in im ϕ, see Theorem 5.3.1(i). By Theorem 5.1.1, εϕ(x) ≥ r for all x, so En(ϕ) = X for n ≤ r, in particular En(ϕ) is closed in this case. Let n > r. By −1 Theorem 5.3.1, En(ϕ) ⊂ X \ ϕ (U). Let Wi be the irreducible compo- −1 nents of the set Y \ U, Wij be the irreducible components of ϕ (Wi) and ϕij : Zij → Wi be the restriction of ϕ. Since dim Wi < dim Y , the set En(ϕij) is closed in Zij, and hence in X. But for n > r we have En(ϕ) = ∪i,jEn(ϕij). 5.4 Open and birational morphisms 69 5.4 Open and birational morphisms Example 5.1.2 shows that the image of an open set under a morphism does not have to be open.

Theorem 5.4.1 Let ϕ : X → Y be a dominant morphism of irreducible varieties, and r = dim X − dim Y . Assume that for each closed irreducible subset W ⊂ Y all irreducible components of ϕ−1(W ) have dimension r + dim W . Then ϕ is open.

Proof Let y ∈ Y . By assumption, all irreducible components of ϕ−1(y) have dimension r. In particular, ϕ−1(y) 6= ∅, whence ϕ is surjective. Moreover, let W ⊂ Y be a closed irreducible subset and Z be an irreducible component of ϕ−1(W ). By assumption, dim Z = r + dim W . Note that ϕ(Z) = W , as otherwise dim ϕ(Z) < dim W , and Z is an irreducible component of ϕ−1(ϕ(Z)), so we get a contradiction with our assumptions. Now, let U be an open subset of X, V = ϕ(U), and y ∈ V . Then y = ϕ(x) for some x ∈ U. It suﬃces to prove that y is in the interior of V . Otherwise y ∈ Y \ V . By Theorem 5.3.3, V is constructible, so Y \ V is also constructible. It follows that y lies in the closure of some locally closed subset O ∩ C contained in Y \ V , where O is open and C is closed. We may assume that C = O ∩ C. Moreover, we may assume that C is irreducible, so O ∩ C is dense in C. Now, each of the irreducible components of the set C0 := ϕ−1(C) dominates C. So the set O0 := ϕ−1(O) intersects each of the components non-trivially. So O0 ∩C0 is dense in C0. But the set O0 ∩C0 = ϕ−1(O∩C) lies in a closed subset X \ U, whence C0 ⊂ X \ U. This contradicts the fact that x ∈ C0.

Irreducible varieties X and Y are called birationally isomorphic, if k(X) is k-isomorphic to k(Y ). A birationally isomorphic varieties do not have to be isomorphic, for example A1 is birationally isomorphic to P1. On the other hand:

Proposition 5.4.2 Let X and Y be irreducible varieties. Then X and Y are birationally isomorphic if and only if there exist non-empty open subsets U ⊂ Y and V ⊂ X which are isomorphic.

Proof The ‘if-part’ is clear. In the other direction, let ϕ : k(Y ) → k(X) be a k-isomorphism. We may assume that X and Y are aﬃne. Let 70 Morphisms f1, . . . , fn generate the ring k[X] over k. Then for each i we can write ϕ(gi) fi = ϕ(h) (gi, h ∈ k[Y ]). So ϕ induces an isomorphism k[Y ]h →˜ k[X]ϕ(h). So we may take U = Yh and V = Xϕ(h).

A bijective morphism does not have to be an isomorphism. In fact its topological behavior and the eﬀect of its comorphism on functions can be quite subtle. A typical example is the Frobenius map Fr : A1 → A1. But even in characteristic 0 one cannot assert that a bijective map is an isomorphism, see Problem 5.5.4. However, Zariski’s Main Theorem claims that a bijective birational morphism ϕ : X → Y of irreducible varieties has to be an isomorphism if Y is smooth. (The smoothness will be deﬁned in the next chapter. We will not prove Zariski’s theorem).

Theorem 5.4.3 Let ϕ : X → Y be a dominant, injective morphism of irreducible varieties. Then k(X) is a ﬁnite purely inseparable extension of ϕ∗k(Y ).

Proof See Humhreys, Theorem 4.6.

5.5 Problems

Problem 5.5.1 Give an example of a constructible set which is not locally closed.

Problem 5.5.2 Prove that the following are equivalent descriptions of the constructible sets in a topological space X: (i) Constructible sets are ﬁnite disjoint union of locally closed sets. (ii) Constructible sets are the sets expressible as

X \ (X2 \ (X3 \···\ Xn)) ... )

for a nested sequence X1 ⊃ X2 ⊃ X3 ⊃ · · · ⊃ Xn of closed subsets. (ii) The class of constructible sets of X is the smallest class including open subsets and closed under the operations of ﬁnite intersec- tions and complementation.

Problem 5.5.3 Prove that a constructible subset of a variety contains a dense open subset of its closure. 5.5 Problems 71

Problem 5.5.4 Deﬁne a morphism ϕ : A1 → A2 by ϕ(x) = (x2, x3). Then X := im ϕ is closed in A2 and the morphism ϕ : A1 → X is bijective, birational and homeomorphism, but it is not an isomorphism. 6 Tangent spaces

In this chapter, unless otherwise stated all varieties are assumed to be irreducible.

6.1 Definition of tangent space 2 If X is the curve f(T1,T2) = 0 in A then our ‘multivariable calculus intuition’ tells us the tangent space to X at x = (x1, x2) ∈ X is the set of solutions of the linear equation ∂f ∂f (x)(T1 − x1) + (x)(T2 − x2) = 0. ∂T1 ∂T2 This ‘tangent space’ is a line unless both partial derivatives are zero at x. More generally, if f ∈ k[T1,...,Tn] set n X ∂f d f = (x)(T − x ). x ∂T i i i=1 i Now, if X ⊂ An is a closed subset and I = I(X) we define the geometric n tangent space Tan(X)x to X at x to be the linear variety Z(J) ⊂ A where the ideal J is generated by all dxf for f ∈ I. We consider Tan(X)x as a verctor space with the origin at x. Problem 6.8.1 is handy for explicit calculations of geometric tangent spaces. For any f(T ) ∈ k[T ], dxf can be considered as a linear function on n A with the origin at x, so on restriction to Tan(X)x, dxf is a linear function on Tan(X)x. By definition, dxf = 0 on Tan(X)x for f ∈ I(X), so we can define the linear function dxf on Tan(X)x for f ∈ k[X]. Thus ∗ dx becomes a linear map from k[X] to Tan(X)x. It is surjective, as ∗ n any g ∈ Tan(X)x is the restriction of a linear polynomial f on A (as usual, origin at x), and dxf = f. Let M be the maximal ideal of k[X]

72 6.1 Definition of tangent space 73 corresponding to x. As k[X] = k⊕M, and dx maps constants to zero, dx ∗ induces a surjective map from M to Tan(X)x. We claim that the kernel 2 2 of this map is M . By the product rule, M ⊆ ker dx. Conversely, let f ∈ M and dxf = 0 on Tan(X)x. Assume that f is the image of some polynomial function f(T ) on An. By Problem 6.8.1 and linear algebra, P we have dxf = i aidxfi for some ai ∈ k and fi ∈ I(X). Then for P n g := f − i aifi we have dxg = 0 on A , which means that g does not contain linear terms (Ti − xi), i.e. g belongs to the square of the ideal generated by all Ti − xi. The image of this ideal in k[X] is M, and the image of g is f, so f ∈ M 2. ∗ 2 Thus, we have identified the vector space Tan(X)x with M/M or 2 ∗ Tan(X)x with (M/M ) . Now, in view of Lemma 2.1.11, the vector 2 2 space M/M can be identified with mx/mx, where mx is the maximal ideal of the local ring Ox. So, we have motivated the following ‘invariant’ definition.

Deﬁnition 6.1.1 The tangent space to the variety X at x ∈ X, denoted 2 ∗ TxX is the k-vector space (mx/mx) .

We now give another description of TxX.A derivation at x is a k- linear map δ : Ox → k such that δ(fg) = δ(f)g(x) + f(x)δ(g). We claim that the vector space of derivations at x is naturally isomorphic to TxX. Indeed, if δ : Ox → k is a derivation, it follows easily that δ(f) = 0 if f 2 2 ∗ is a constant or if f ∈ mx. So δ defines an element of (mx/mx) . This defines a map from the space of derivations at x to TxX, which is easily shown to be an isomorphism. Let X be an irreducible affine variety. We claim that in this case we can also identify TxX with the derivations of k[X] at x, i.e. the linear maps δ : k[X] → k such that δ(fg) = δ(f)g(x) + f(x)δ(g). Indeed, recall that under our assumptions Ox can be identified with the subring of k(X) consisting of all rational functions which are regular at x. Now, ¯ if δ : Ox → k is a derivation, we get a derivation δ : k[X] → k on f restriction. Conversely, if δ : k[X] → k is a derivation and h = g ∈ Ox ˆ δ(f)g(x)−f(x)δ(g) define δ(h) = g(x)2 (the ‘quotient rule’). It is easy to check that the maps δ 7→ δ¯ and δ 7→ δˆ are inverse to each other.

n ∂ ∂f(x) Example 6.1.2 Let X = . The map |x : k[X] → k, f 7→ A ∂Ti ∂Ti is a derivation of k[X] = k[T1,...,Tn] at x. It is easy to check that ∂ ∂ ∼ n the derivations |x,..., |x form a basis of TxX, so TxX = k . It ∂T1 ∂Tn n ∼ n n follows that TxP = k for any x ∈ P . 74 Tangent spaces

Example 6.1.3 Let X ⊂ An be an aﬃne irreducible variety. Then any derivation δ of k[X] = k[T1,...,Tn]/I(X) can be lifted to a derivation δˆ of k[X] at x. So by Example 6.1.2 any derivation of X at Pn ∂ x looks like ai |x for some constants ai ∈ k. Moreover, if i=1 ∂Ti Pn ∂ I(X) = (f1, . . . , fl), then ai |x is zero on I(X) if and only if i=1 ∂Ti n X ∂fj(x) a (j = 1, . . . , l). (6.1) i ∂T i=1 i

n So TxX is a linear space of all tuples (a1, . . . , an) ∈ k satisfying the equations (6.1).

Example 6.1.4 Let X be given by the equation y2 = x3. Then I(X) = (y2 − x3), and X is one-dimensional. On the other hand, using Example 6.1.3, one sees that the tangent space TxX is one-dimensional for all x, except for x = (0, 0) when TxX is two-dimensional. Soon we will see that in general dim TxX ≥ dim X, and that the equality holds for ‘almost all’ points x ∈ X.

Proposition 6.1.5 Let X,Y be irreducible varieties, x ∈ X, y ∈ Y . ∼ Then T(x,y)(X × Y ) = TxX ⊕ TyY .

Proof We may assume that X and Y are aﬃne. If δ1 : k[X] → k is a derivation at x and δ2 : k[Y ] → k is a derivation of k[Y ] at y, deﬁne the derivation

(δ1, δ2): k[X × Y ] = k[X] ⊗ k[Y ] → k, f ⊗ g 7→ δ1(f)g(y) + f(x)δ2(y) of k[X × Y ] at (x, y). This deﬁnes an isomorphism from TxX ⊕ TyY to T(x,y)(X × Y ) (check!).

6.2 Simple points

Deﬁnition 6.2.1 Let X be an irreducible variety and x ∈ X. Then x is called a simple point if dim TxX = dim X. Otherwise x is called singular. If all points of X are simple, then X is called smooth (or non-singular).

So An and Pn are smooth and the product of smooth varieties is smooth. 6.2 Simple points 75

Lemma 6.2.2 Let X = Z(f) be an irreducible hypersurface in An. Then dim TxX = dim X for all points x from some open dense subset of X.

Proof We may assume that f is an irreducible polynomial. The tangent n space TxX is the set of all n-tuples (a1, . . . , an) ∈ k satisfying the linear Pn ∂f(x) equation ai = 0. Since dim X = n − 1, a point x is singular i=1 ∂Ti if and only if all ∂f(x) = 0. If the polynomial ∂f is non-zero, then it is ∂Ti ∂Ti not identically zero on X, as otherwise it would be divisible by f, which is impossible by degrees. So we may assume that char k = p and all p degrees of all variables Ti in f are divisible by p, but then f = g by ‘Freshman’s Dream’, which contradicts the irreducibility of f.

Theorem 6.2.3 Let X be an irreducible variety. Then dim TxX ≥ dim X for any x ∈ X, and the equality holds for all points x from some open dense subset of X.

Proof By Theorem 2.1.12, k(X) is separably generated over k, i.e. k(X) is a finite separable extension of L = k(t1, . . . , td), which in turn is a purely transcendental extension of k. Note that d = dim X. By the Primitive Element Theorem 2.1.13, K = L(t0) for some element t0 ∈ K. Let f(T0) := irr (t0; L) ∈ L[T0] be the minimal polynomial. Since the coefficients of f are rational functions in k(t1, . . . , td), this polynomial can be considered as a rational function f(T0,T1,...,Td) ∈ k(T0,T1,...,Td). This rational function is defined on a principal open subset of Ad+1, and the zero locus Y of f is an irreducible hypersurface in this principal open subset. ∼ We claim that k(Y ) = k(X). Indeed, let si be the restriction of the coordinate function Ti to Y for 0 ≤ i ≤ n. Then k(Y ) = k(s0, s1, . . . , sd). As dim Y = d and s0 is algebraic over k(s1, . . . , sd), we conclude that s1, . . . , sd are algebraically independent over k. Now, it is clear that the minimal polynomial of s0 over k(s1, . . . , sd) is f, whence the claim. By Proposition 5.4.2, there exist non-empty open subsets in X and Y which are isomorphic. By Lemma 6.2.2, the set of points y ∈ Y for which dim TyY = dim Y form an open subset in Y , so the same follows for X.

Let x be an arbitrary point of X. In order to find the dimension of TxX we may pass to an affine open neighborhood of x. So we may assume that n X is a closed subset of some A . Then TxX can be considered as a vector n subspace of k . By shifting the origin to x we have TxX as an affine 76 Tangent spaces subspace of An through x. Let T be the subset of all (x, y) ∈ X ×An for which y ∈ TxX. Note that T is a closed subset. Indeed, it is given by the equations for X together with the polynomial equations of the form Pn ∂fj (x) n (Si − xi), where Si are the coordinates in . Projection i=1 ∂Ti A −1 pr1 defines a morphism ϕ : T → X whose fiber ϕ (x) has dimension −1 dim TxX. For each m the subset Xm = {x ∈ X | dim ϕ (x) ≥ m} is closed in X, see Proposition 5.3.4(i). But we saw that Xd is dense in X, so Xd = X.

6.3 Local ring of a simple point Let X be an irreducible variety and x ∈ X. By Corollary 2.1.10, the minimal number n of generators of the ideal mx equals the dimension of 2 mx/mx or dim TxX. So the point x is simple if and only if n = dim X. Recall that the Krull dimension of a Noetherian ring R is deﬁned to be the largest length k of a chain

0 ( P1 ( P2 ( ··· ( Pk ( R of prime ideals. We claim that the Krull dimension of Ox equals dim X. Indeed, we may assume that X is aﬃne, in which case Ox = k[X]Mx . But the prime ideals of k[X]Mx are in one-to-one correspondence with prime ideals of k[X] contained in Mx, and it follows from Corollaries 4.5.11 and 4.5.12 that dim X is the largest length k of a chain

0 ( P1 ( P2 ( ··· ( Pk = Mx of prime ideals in k[X]. A local ring (R,M) is called regular if its Krull dimension equals the number of generators of the maximal ideal M. We have established that a point x ∈ X is simple if and only if its local ring Ox is regular. So, in view of Theorem 2.1.15, we have:

Theorem 6.3.1 Let x be a simple point of an irreducible variety X. Then Ox is a regular local ring. In particular, it is a UFD and is integrally closed.

Theorem 6.3.2 Let X be an irreducible variety and x ∈ X be a point such that Ox is integrally closed. Let f ∈ K(X) \Ox. Then there exists 0 1 a subvariety Y ⊂ X containing x and such that f := f ∈ Oy for some y ∈ Y , and f 0 is equal to zero on Y everywhere where it is deﬁned. 6.4 Diﬀerential of a morphism 77

Proof Let R = Ox. Then I := {g ∈ R | gf ∈ R} is a proper ideal of R, as 1 6∈ I, and so I ⊂ mx. Let P = P1,P2,...,Pt be the distinct minimal prime ideals containing I. Then P1 ∩ · · · ∩ Pt/I is nilpotent, i.e. n n n P P2 ...Pt ⊂ I. For i > 1, Pi generates in the local ring RP ⊂ k(X) n the ideal coinciding with the whole RP . So P RP ⊂ IRP . In particular, n since If ⊂ R, we have P f ⊂ (If)Rp ⊂ RP . Choose k ≥ 0 the minimal k k−1 possible so that P f ⊂ RP , and let g ∈ P f \ RP . Then P g ⊂ RP . By assumption R is integrally closed, so RP is integrally closed, see Proposition 2.2.13. As g 6∈ RP , the element g is not integral over RP . Now, if PRP g ⊆ PRP , then the ring RP [g] acts faithfully on the finitely generated RP -module PRP , giving a contradiction, see Proposi- tion 2.2.5(iii). So P g ⊂ RP generates the ideal RP in RP , hence contains 1 1 an invertible element from RP . So g ∈ PRP , and PRP = g RP . f k Now, h := gk ∈ fP RP ⊂ RP . We claim that h is a unit in RP . 1 f Otherwise h ∈ PRP = g RP or gk−1 ∈ RP , which contradicts the choice 1 −1 1 of k. So f = h gk ∈ PRP . Let P be generated by the elements f1, . . . , fl ∈ Ox. Then fi are rational functions regular in some neighborhood of x, so also in some affine neighborhood of x. Now let Y be the zero locus of the functions f1, . . . , fl in this affine neighborhood. Then all functions of P are zero everywhere on Y where they are defined. So this is also true for the 1 function f ∈ PRP . Also x ∈ Y , as P ⊂ mx.

6.4 Diﬀerential of a morphism Let ϕ : X → Y be a morphism of (irreducible) varieties, x ∈ X, y = ∗ ∗ ∗ ϕ(x). Then ϕ (Oy) ⊂ Ox, and ϕ (my) ⊂ mx. So ϕ induces a map 2 2 2 ∗ 2 ∗ my/my → mx/mx, which in turn induces a map (mx/mx) → (my/my) . This map is denoted dϕx and is called the diﬀerential of ϕ at x. Thus:

dϕx : TxX → Tϕ(x)Y.

In terms of derivations, dϕx can be described similarly: if δ : Ox → k is ∗ a derivation, then dϕx(δ) is deﬁned to be δ ◦ϕ : Oy → k. The following natural properties are easy to check:

dx id = id and d(ψ ◦ ϕ)x = dψϕ(x) ◦ dϕx.

Example 6.4.1 Let X ⊂ An and Y ⊂ Am be aﬃne algebraic sets and ϕ : X → Y be the restriction of ϕ = (ϕ1, . . . , ϕm) with ϕi ∈ k[T1,...,Tn]. Take x ∈ X and let y = ϕ(x). We identify TxX and TyY 78 Tangent spaces with subspaces of kn and km, respectively, following Example 6.1.3. If a = (a1, . . . , an) ∈ TxX, then dϕx(a) = (b1, . . . , bm), where

X ∂ϕj b = (x)a , j ∂T i i i i.e. dϕx is the linear map whose matrix is the Jacobian of ϕ at x.

× Example 6.4.2 Let X = GLn(k), Y = GL1(k) = k , and ϕ = det. 2 Note that X is the principal open set in An which we identify with Mn(k), all n×n matrices. It is easy to see that at every point x ∈ GLn(k) the tangent space TxGLn(k) can be identiﬁed with Mn(k). Let e be the identity matrix. Under our identiﬁcation, d dete : Mn(k) → M1(k) = k is tr, the trace map.

Let ϕ : X → Y be a dominant morphism of irreducible varieties. Then k(Y ) can be considered as a subﬁeld of k(X) via ϕ∗. If the extension k(X)/k(Y ) is separable, we say that the morphism ϕ is separable. In characteristic 0 all morphisms are separable. An example of a non- separable morphism is given by the Frobenius morphism. We are going to develop some machinery which will help us to establish a diﬀerential criterion for separability and to consider tangent spaces from a new point of view.

6.5 Module of diﬀerentials For a k-algebra A and an A-module M, we write

Derk(A, M) for the space of all k-linear derivations from A to M, i.e. k-linear maps f : A → M such that f(ab) = af(b) + bf(a) for all a, b ∈ A.

Let m : A ⊗k A → A be the multiplication, and let I := ker m, the ideal generated by all a ⊗ 1 − 1 ⊗ a (a ∈ A). Deﬁne the module of diﬀerentials ΩA/k to be 2 ΩA/k := I/I .

This is an A ⊗k A-module annihilated by I, so it can be considered as a module over A =∼ (A ⊗ A)/I. Let da denote the image of a ⊗ 1 − 1 ⊗ a in ΩA/k. Note that the map d : a 7→ da is a derivation from A to the 6.5 Module of diﬀerentials 79

A-module ΩA/k: ad(b) + d(a)b = a(b ⊗ 1 − 1 ⊗ b) + (a ⊗ 1 − 1 ⊗ a)b + I2 = ab ⊗ 1 − 1 ⊗ ab + I2 = d(ab).

The elements da for a ∈ A generate ΩA/k as an A-module. One should think of ΩA/k as the universal module for derivations of A:

Theorem 6.5.1 Suppose that M is an A-module and D : A → M is a k-derivation. Then there exists a unique A-module homomorphism

ϕ :ΩA/k → M such that D = ϕ ◦ d, i.e. the map

HomA(ΩA/k,M) → Derk(A, M), ϕ 7→ ϕ ◦ d is an isomorphism.

Proof Deﬁne the linear map

ψ : A ⊗ A → M, a ⊗ b 7→ bD(a).

One checks that for arbitrary elements x, y ∈ A ⊗ A,

ψ(xy) = m(x)ψ(y) + m(y)ψ(x),

2 hence ψ vanishes on I . Therefore it induces a map ϕ :ΩA/k → M which is actually an A-module map, such that ϕ(da) = ψ(a⊗1−1⊗a) = D(a) (here we have used that D(1) = 0). For uniqueness use the fact that the da generate ΩA/k as an A-module.

The theorem gives a universal property for the pair (ΩA/k, d) which as usual characterizes it up to a unique A-module isomorphism.

Example 6.5.2 (i) Let F be any ﬁeld, A = F [T1,...,Tn]/(f1, . . . , fm), and ti = Ti + (f1, . . . , fm) ∈ A. Then the dti generate ΩA/F as an A- module, since the ti generate A as an algebra. Moreover, the kernel of the A-module homomorphism

n n M A = Aei → ΩA/F , ei 7→ dti i=1 is the submodule K of An generated by the elements

n X ∂fj (t , . . . , t )e (1 ≤ j ≤ m). ∂T 1 n i i=1 i 80 Tangent spaces Indeed, consider the map n X ∂f d0 : A → An/K, f 7→ (t , . . . , t )e ∂T 1 n i i=1 i ∂f ˜ ( (t1, . . . , tn) means: take any representative f(T1,...,Tn) in F [T ], ∂Ti take the partial derivative of f˜, and pass to the quotient again.) The result follows from the fact that (An/K, d0) satisfy the universal property of the theorem.

(ii) Consider two special cases of (i): when A = k[T1,...,Tn], then 2 ΩA/k is a free module on the basis dT1, . . . , dTn; when A = k[T1,T2]/(T1 − 3 2 T2 ), then ΩA/k = (Ae1 ⊕ Ae2)/(2t1e1 − 3t2e2), which is not a free A- module.

(iii) Let A be an integral domain with quotient field E. Then ΩE/k = E ⊗A ΩA/k. Indeed, the derivation d : A → ΩA/k induces a derivation ˆ ˆ d : E → E ⊗A ΩA/k. We claim that E ⊗A ΩA/k together with d has the correct universal property. Take an E-module M and a derivation Dˆ : E → M. Its restriction D to A is a derivation A → M. Hence there exists a unique A-module homomorphism ϕ :ΩA/k → M with D = ϕ ◦ d. Hence since M is an E-module, there is a unique E-module ˆ ˆ homomorphismϕ ˆ : E ⊗A ΩA/k → M with D =ϕ ˆ ◦ d. (iv) Suppose that E = k(x1, . . . , xn) is a finitely generated field extension of k. By (iii) and (i), ΩE/k is the E-vector space spanned by dx1, . . . , dxn. In particular, it is finite dimensional.

Example 6.5.3 Let X be an aﬃne variety, x ∈ X, and kx = k be the 1-dimensional k[X]-module with action f · c = f(x)c for f ∈ k[X], c ∈ k.

Denote ΩX := Ωk[X]/k. By the theorem, ∼ ∼ Homk[X](ΩX , kx) = Derk(k[X], kx) = TxX

n Now, if X ⊂ A is closed and I(X) = (f1, . . . , fm), it follows from Example 6.5.2(i) that ΩX is generated by dt1, . . . , dtm and the relations n X ∂fj (t , . . . , t )dt = 0 (1 ≤ j ≤ m). ∂T 1 n i i=1 i

Now, it is clear that Homk[X](ΩX , kx) is the vector space of all n-tuples (a1, . . . , an) satisfying equations n X ∂fj (x)a = 0 (1 ≤ j ≤ m). ∂T i i=1 i 6.5 Module of diﬀerentials 81 So we recover the description of the tangent space from Example 6.1.3.

Now for the remainder of the section we will be concerned with the following situation: we are given ﬁnitely generated ﬁeld extensions F/E/k. Then there exists an exact sequence

0 → DerE(F,F ) → Derk(F,F ) → Derk(E,F ). The ﬁrst map is the obvious inclusion and the second map is induced by restriction of functions from F to E. To check the exactness in the second term, note that any D ∈ DerE(F,F ) maps E to zero, and conversely, any f ∈ Derk(F,F ) that maps elements of E to zero is E-linear. Applying the universal property we have an exact sequence

0 → HomF (ΩF/E,F ) → HomF (ΩF/k,F ) → HomE(ΩE/k,F ). ∼ Also note that HomE(ΩE/k,F ) = HomF (F ⊗ ΩE/k,F ). So we have an exact sequence of ﬁnite dimensional F -vector spaces

0 → HomF (ΩF/E,F ) → HomF (ΩF/k,F ) → HomF (F ⊗ ΩE/k,F ). Dualizing we get

α β F ⊗ ΩE/k −→ ΩF/k −→ ΩF/E −→ 0, where α sends 1 ⊗ dE/ka to dF/ka, viewing a ∈ E as an element of F , and β is induced by the derivation dF/E : F → ΩF/E according to the universal property of ΩF/k.

Lemma 6.5.4 If F is a ﬁnite dimensional separable extension of E then α is injective.

Proof By the above discussion, this is equivalent to the restriction map Derk(F,F ) → Derk(E,F ) being surjective. Equivalently, every k- derivation from E to F can be extended to a derivation from F to F . By the Primitive Element Theorem, we may assume that F = E[T ]/(f(T )), where n n−1 f(T ) = T + an−1T + ··· + a0 is an irreducible polynomial and (this is what separability means) f 0(x) 6= 0, where x is the image under the quotient map of T in F . Let D : E → F be a derivation. To extend D to a derivation Dˆ from F to F , we just need to decide what Dˆ(x) should be: then the derivation formula means that there is no choice for deﬁning Dˆ applied to any other 82 Tangent spaces element of F = E[x]. To decide on Dˆ(x) we need for well-deﬁnedness that Dˆ(f(x)) = 0, i.e.

0 X i f (x)Dˆ(x) + D(ai)x = 0.

Since f 0(x) 6= 0, we can solve this equation for Dˆ(x) in the ﬁeld F .

Lemma 6.5.5 Let F = E(x). Then dimF ΩF/E ≤ 1. Moreover, ΩF/E = 0 if and only if F/E is a ﬁnite separable extension.

Proof By Example 6.5.2(iii), ΩF/E = F ⊗E[x] ΩE[x]/E. If x is transcendental over E, we have ΩE[x]/E is free of rank 1, cf. Example 6.5.2(i). If x 0 is algebraic, by Example 6.5.2(i) again we have ΩE[x]/E = E[x]/(f (x)), where f(x) = irr (x; E). If F/E is not separable, then f 0(x) = 0, and we again get that ΩF/E is one-dimensional. Finally, if F/E is separable, 0 then f (x) 6= 0, and ΩF/E = F ⊗E[x] ΩE[x]/E = 0, since it is generated by 1 ⊗ 1 = f 0(x)−1f 0(x) ⊗ 1 = 0.

Theorem 6.5.6 (Differential Criterion for Separability) Let F = E(x1, . . . , xm) be a finitely generated field extension. Then:

(i) dimF ΩF/E ≥ tr. degE F . (ii) Equality in (i) holds if and only if F/E is a separable extension.

Proof Proceed by induction on d = dimF ΩF/E. If d = 0, i.e. ΩF/E = 0 to get (i) and (ii), we just need to show that F/E is a ﬁnite separable extension. For this we use induction on m, the case m = 1 be- 0 ing Lemma 6.5.5. Now suppose m > 1. Set E = E(xm), so F = 0 E (x1, . . . , xm−1). Using the exact sequence

α β F ⊗ ΩE0/E −→ ΩF/E −→ ΩF/E0 −→ 0,

0 we see that ΩF/E0 = 0. Hence by induction F/E is a finite separable extension. So by Lemma 6.5.4, α is injective, whence ΩE0/E = 0, and E0/E is a finite separable extension. By transitivity, F/E is a finite separable extension. 0 Now suppose d > 0. Pick x ∈ F with dF/Ex 6= 0, and let E := E(x). We have the exact sequence

α β F ⊗ ΩE0/E −→ ΩF/E −→ ΩF/E0 −→ 0.

Since α(1⊗dE0/Ex) = dF/Ex 6= 0, we have ΩE0/E 6= 0. So by Lemma 6.5.5, 6.6 Simple points revisited 83 dimE0 ΩE0/E = 1, which means that α is injective. So dimF ΩF/E = dimF ΩF/E0 + 1. By induction, dimF ΩF/E ≥ tr. degE0 F + 1. Since 0 tr. degE F = tr. degE0 F + tr. degE E ≤ tr. degE0 F + 1, we get dimF ΩF/E ≥ tr. degE F , which is (i). With a little further argument along the same lines, one gets (ii).

Corollary 6.5.7 Assume that E ⊂ F are ﬁnitely generated ﬁeld extensions of k. Then F/E is separable if and only if the natural map Derk(F,F ) → Derk(E,F ) is surjective.

Proof As above, Derk(F,F ) → Derk(E,F ) is surjective if and only if the map

α : F ⊗E ΩE/k → ΩF/k is injective. Consider the exact sequence

α β F ⊗ ΩE/k −→ ΩF/k −→ ΩF/E −→ 0. As k is algebraically closed, every extension of k is separable, so by the theorem, dimF F ⊗E ΩE/k = dimE ΩE/k = tr. degk E and dimF ΩF/k = tr. degk F . Hence α is injective if and only if

dimF ΩF/E = tr. degk F − tr. degk E = tr. degE F. By the theorem, this is if and only if F/E is separable.

6.6 Simple points revisited Suppose that A is an integral domain with ﬁeld of fractions F . Let R = (ri,j) be an m×n matrix with entries in A. Consider the A-module n n M X MA(R) := Aei/h ri,jej | i = 1, . . . , mi j=1 j=1 given by generators and relations. If Y is an invertible m × m matrix ∼ with entries in A, then the change of basis argument gives MA(YR) = MA(R). Similarly, if Z is an invertible n × n matrix with entries in A, ∼ then MA(RZ) = MA(R). Now by linear algebra we can ﬁnd invertible matrices Y and Z with entries in F such that I 0 R = Y r Z, 0 0 84 Tangent spaces where r is the rank of R. Putting all entries of Y and Z over a common denominator, we may assume that Y and Z have entries in Af for some non-zero f ∈ A. Note that ∼ MA(R)f = MAf (R).

So MA(R)f is a free Af -module of rank n − r. Recall that if X is an aﬃne variety, we write ΩX for Ωk[X]/k. If x ∈ X, let us also write ΩX (x) for the vector space kx ⊗k[X] ΩX . This is called ∼ ∗ cotangent space for ΩX (x) = (TxX) . Indeed, using Example 6.5.3, we have ∼ ∗ TxX = Homk[X](ΩX , kx) = Homk(kx ⊗k[X] ΩX , k) = ΩX (x) .

∂fj If k[X] = k[T1,...,Tn]/(f1, . . . , fm), let R be the m×n matrix ( (t)) ∂Ti ∂fj and R(x) = ( (x)). Then ΩX = M (R) and ΩX (x) = Mk(R(x)). ∂Ti k[X]

Lemma 6.6.1 Assume that X is an irreducible aﬃne variety.

(i) dimk(X) Mk(X)(R) = dim X. (ii) If x ∈ X is a simple point, then there is f ∈ k[X] with f(x) 6= 0

such that Mk[X](R)f is a free k[X]f -module of rank dim X with basis given by dim X out of the images of the ei.

Proof (i) Since k is algebraically closed, k(X) is a separable extension of k. So Theorem 6.5.6 tells us that dim X = dimk(X) Ωk(X)/k. But ∼ Ωk(X)/k = k(X) ⊗k[X] ΩX = Mk(X)(R). (ii) In view of (i), the rank of the matrix R is r := n − dim X. Some r×r-minor of R(x) has non-zero determinant. Reordering if necessary we may assume that this is the principal minor in the top left hand corner. Let f be the determinant of this minor, so f(x) 6= 0. On localizing at f, the matrix R becomes equivalent to I 0 r . 0 0

The lemma implies

Theorem 6.6.2 Let X be an irreducible variety. If x ∈ X is a simple point, there is an aﬃne neighborhood U of x such that ΩU is a free k[U]-module on basis dg1, . . . , dgdim X for suitable gi ∈ k[U]. 6.7 Separable morphisms 85 6.7 Separable morphisms

Recall that if X is an affine variety, we write ΩX for Ωk[X]/k and ΩX (x) for the cotangent space kx ⊗k[X] ΩX at X. Let ϕ : X → Y be a separable dominant morphism of irreducible affine ∗ varieties. The composition of ϕ : k[Y ] → k[X] and dX : k[X] → ΩX is a derivation ∗ dX ◦ ϕ : k[Y ] → ΩX . So by the universal property of the differentials we get induced a k[Y ]- module map ∗ ϕˆ :ΩY → ΩX

∗ ∗ such that dX ◦ ϕ =ϕ ˆ ◦ dY . Let x ∈ X and y = ϕ(x). The k[X]-module kx viewed as a k[Y ]- ∗ module via ϕ is ky. After identifying TxX with Homk[X](ΩX , kx) and TyY with Homk[Y ](ΩY , ky), the map dϕx becomes:

∗ dϕx : Homk[X](ΩX , kx) → Homk[Y ](ΩY , ky), θ 7→ θ ◦ ϕˆ .

Theorem 6.7.1 Let ϕ : X → Y be a morphism of irreducible varieties. (i) Assume that x ∈ X and y = ϕ(x) ∈ Y are simple points and that dϕx is surjective. Then ϕ is a dominant separable morphism. (ii) Assume that ϕ is a dominant separable morphism. Then the simple points x ∈ X with ϕ(x) simple and dϕx surjective form a non-empty open subset of X.

Proof (i) We may assume that X and Y are aﬃne and ΩX ,ΩY are free K[X]-, resp. k[Y ]-modules of rank d = dim X resp. e = dim Y . ∗ In particular X and Y are smooth. The mapϕ ˆ :ΩY → ΩX of k[Y ]- modules induces a homomorphism of free k[X]-modules

ψ : k[X] ⊗k[Y ] ΩY → ΩX . We can represent ψ as a d×e-matrix A with entries in k[X], ﬁxing bases for ΩX and ΩY . Suppose that dϕx is surjective. Then A(x), which ∗ represents the dual map dϕx :ΩY (y) → ΩX (x), is injective, hence a matrix of rank e. Hence the rank of A itself is at least e, hence equal to e since rank cannot be more than the number of columns. This shows ∗ that ψ is injective. Henceϕ ˆ is injective too. Since ΩX and ΩY are free modules, this implies that ϕ∗ : k[Y ] → k[X] must be injective. So ϕ must be dominant. 86 Tangent spaces Moreover, injectivity of ψ implies the injectivity of

k(X) ⊗k[Y ] ΩY → k(X) ⊗k[X] ΩX .

This is the map α in the exact sequence

α β k(X) ⊗k(Y ) Ωk(Y )/k −→ Ωk(X)/k −→ Ωk(X)/k(Y ) −→ 0.

Hence k(X) is a separable extension of k(Y ) by the diﬀerential criterion for separability.

Example 6.7.2 I will illustarte the usefulness of the theorem by an example from my research. Recently Jon Brundan and I needed to establish the following. Consider the polynomial algebra

[r] C[xij | 1 ≤ i, j ≤ n, r = 1, . . . , l], and let

(r) X X [s ] [s ] [s ] y = x 1 x 2 ··· x r i,j i0,i1 i1,i2 ir−1,ir 1≤s1<···

1≤s1<···

∗ [r] (r) in the matrices A1,...,Al. The comorphism θ maps xi,j to yi,j . So to (r) show that the yi,j are algebraically independent, we need to show that θ∗ is injective, i.e. that θ is a dominant morphism of affine varieties. For this it suffices to show that the differential of θ is surjective at some ×l point x ∈ Mn . Pick pairwise distinct scalars c1, . . . , cl ∈ C and consider

x := (c1In, . . . , clIn). 6.8 Problems 87

×l ⊕l Identifying the tangent space Tx(Mn ) with the vector space Mn , a calculation shows that the diﬀerential dθx maps (A1,...,Al) to (B1,...,Bl) where l X Br = er−1(c1,..., cbs, . . . , cl)As. s=1

Here er−1(c1,..., cbs, . . . , cl) denotes the (r − 1)th elementary symmetric function in the scalars c1, . . . , cl excluding cs. We just need to show this linear map is surjective, for which it clearly suﬃces to consider the case n = 1. But in that case its determinant is the Vandermonde determinant Q 1≤r

6.8 Problems

Problem 6.8.1 Let X ⊂ An be a closed subset, I = I(X), and J be the ideal of k[T1,...,Tn] generated by all dxf for f ∈ I. If f1, . . . , fl generate I, then dxf1, . . . , dxfl generate J. 7 Complete Varieties

7.1 Main Properties

A variety X is called complete if for any variety Y the projection π2 : X × Y → Y is a closed map.

Remark 7.1.1 Completeness is an algebraic analogue of compactness. To be more precise, let X be a locally compact Hausdorﬀ topological space. One can prove that X is compact if and only if for any locally compact space Y the projection π2 : X × Y → Y is closed.

Example 7.1.2

(i) A point is complete, as if X is a point, π2 : X × Y → Y is an isomorphism. 1 1 1 (ii) A is not complete. Indeed, take Z = Z(T1T2 − 1) ⊂ A × A = 2 1 A . Then π2 maps Z onto A \{0}.

Remark 7.1.3 (i) X is complete if and only if all its irreducible components are complete. (ii) X is complete if for any irreducible aﬃne variety Y the projection π2 : X × Y → Y is closed.

Proposition 7.1.4 Let X,Y be varieties. (i) If X is complete and Y ⊂ X is closed then Y is complete. (ii) If X and Y are complete, then so is X × Y . (iii) If ϕ : X → Y is a morphism and X is complete, then ϕ(X) is closed and complete. (iv) If Y is a complete subvariety of X, then Y is closed.

88 7.2 Completeness of projective varieties 89

(v) If X is complete and irreducible, then OX (X) = k. In particular, if X is complete and aﬃne, then X is a ﬁnite number of points.

Proof (i) A closed subset of Y × Z is also closed in X × Z.

(ii) Projection X × Y × Z is a composition of πY × idZ : X × Y × Z → Y × Z and πZ : Y × Z → Z. (iii) Since Y is a variety, the graph of ϕ is closed in X × Y . Its image is ϕ(X), which is closed by completeness of X. To show completeness of ϕ(X), take a closed subset K ⊂ ϕ(X) × Z for some Z. Consider 0 projections π2 : X × Z → Z, π2 : ϕ(X) × Z → Z, and note that 0 −1 π2(K) = π2((ϕ × idZ ) (K)). (iv) Apply (iii) to the embedding of Y into X. 1 (v) Let f ∈ OX (X). Then f is a morphism from X to A , cf. Prob- lem 4.6.2. By (iii), f(X) is closed complete irreducible subvariety of A1, and it could not be A1 itself, since A1 is not complete, so f(X) is a point, i.e. f is a constant.

7.2 Completeness of projective varieties

Theorem 7.2.1 Any projective variety is complete.

Proof In view of Proposition 7.1.4(i) and Remark 7.1.3(ii), it suffices to n prove that π2 : P × Y → Y is closed for any irreducible affine variety Y . Set R := k[Y ]. n n For 0 ≤ i ≤ n, let Pi be the affine open set of P given by Xi 6= 0, n where X0,X1,...,Xn are the coordinate ‘functions’ on P . Then the n n affine open sets Ui := Pi × Y cover P × Y . Moreover, we can identify k[Ui] with Ri := k[X0/Xi,...,Xn/Xi] ⊗ R = R[X0/Xi,...,Xn/Xi]. n Let Z be any cosed set in P × Y , and y ∈ Y \ π2(Z). We want to find a neighborhood of y in Y of the form Yf which is disjoint from π2(Z). This amounts to finding f ∈ R with f 6∈ M := My and such that f i vanishes on π2(Z). Let π2 := π2|Ui and Zi := Z ∩ Ui, 0 ≤ i ≤ n. Now i ∗ f|π2(Z) ≡ 0 is equivalent to the statement that the pullback of (π2) (f) i ∗ is zero on Zi ⇔ (π2) (f) ∈ I(Zi) /Ri. The existence of such f will follow from Nakayama’s Lemma applied to a suitable R-module, which we now construct.

First consider the polynomial ring S := R[X0,...,Xn] with natural grading S = ⊕mSm. We construct the homogeneous ideal I/S by letting 90 Complete Varieties

Im consist of all f(X0,...,Xn) ∈ Sm such that f(X0/Xi,...,Xn/Xi) ∈ I(Zi) for each i. Next, fix i and let f ∈ I(Zi). We claim that the multiplication of f by a sufficiently high power of Xi will take f into I. Indeed, for large m m, Xi f becomes a homogeneous polynomial of degree m. Moreover, m m m+1 m+1 (Xi /Xj )f ∈ Rj vanishes on Zi ∩ Uj = Zj ∩ Ui, while (Xi /Xj )f m+1 m+1 vanishes at all points of Zj not in Ui. So (Xi /Xj )f vanishes on m+1 Zj. Since j is arbitrary, we conclude that Xi f lies in Im+1. n Now, Zi and Pi × {y} are disjoint closed subsets of the affine variety Ui, so their ideals I(Zi) and MRi generate Ri, i.e. we can write 1 = P fi + j mijgij, where fi ∈ I(Zi), mij ∈ M, and gij ∈ Ri. By the preceding paragraph, multiplication by a sufficiently high power of Xi takes fi into I. We can choose this power large enough to work in these equations for all fi and to take gij into S as well. So we obtain m Xi ∈ Im + MSm for all i. Enlarging m even more, we can get all monomials of degree m in X0,...,Xn to lie in Im + MSm. This implies Sm = Im + MSm. Now apply Corollary 2.1.9 to the finitely generated R-module Sm/Im, which satisfies M(Sm/Im) = Sm/Im. The conclusion is that there exists f ∈ R\M such that f annihilates Sm/Im, i.e. fSm ⊂ Im. In particular, m m fXi ∈ Im, so by definition of Im we have (fXi )(X0/Xi,...,Xn/Xi) ∈ m I(Zi), but (fXi )(X0/Xi,...,Xn/Xi) ∈ I(Zi) = f. Part two Algebraic Groups

8 Basic Concepts

8.1 Deﬁnition and ﬁrst examples

Deﬁnition 8.1.1 An algebraic group is an aﬃne variety G equipped with morphisms of varieties µ : G × G → G, ι : G → G that give G the structure of a group. A morphism f : G → H of algebraic groups is a morphism of varieties that is a group homomorphism too.

It is possible to consider algebraic groups which are not necessarily affine varieties, so strictly speaking one we should have used the term affine algebraic group above. As we will only meet affine algebraic groups we will drop the word ‘affine’. The kernel of a morphism f : G → H of algebraic groups is a closed subgroup of G, so it is an algebraic group in its own right. The same will turn out to be true about the images. Translation by an element g ∈ G is an isomorphism of varieties, so all geometric properties at one point can be transferred to any other point. For example, as G has simple points, G is smooth.

Example 8.1.2 (i) The additive group Ga is the group (k, +), i.e. the aﬃne variety A1 under addition. × (ii) The multiplicative group Gm is the group (k , ×), i.e. the principal open subset A1 \{0} under multiplication. (iii) The group GLn = GLn(k) is the group of all invertible n × n ma- n2 trices over k. As a variety, this is a principal open set in Mn(k) = A corresponding to the determinant. Since the formulas for matrix multiplication and inversion involve only polynomials in the matrix entries and 1/ det, the group structure maps are morphisms of varieties. Let V be an n-dimensional vector space over k. Then by ﬁxing a

93 94 Basic Concepts basis we can deﬁne a structure of an algebraic group on GL(V ) which ∼ is independent of the choice of basis. Of course, GL(V ) = GLn.

(iv) The group SLn = SLn(k) is the closed subgroup of GLn deﬁned by the zeros of det −1.

(v) The group Dn of invertible diagonal matrices is a closed subgroup of GLn (given be the zeros of which functions?) It is isomorphic to the direct product Gm × · · · × Gm (m copies). (vi) The group Un of upper unitriangular matrices is another closed subgroup of GLn. t (vii) The orthogonal group On = {x ∈ GLn | xx = 1}. We exclude the characteristic 2 when considering this example...

(viii) The special orthogonal group SOn = On ∩ SLn is a normal subgroup of On of index 2. t (ix) The symplectic group Sp2n = {x ∈ GLn | x Jx = J} where 0 In −In 0 is another closed subgroup.

Let G be an (aﬃne) algebraic group with the identity element e, and put A = k[G]. The map

ε : A → k, f 7→ f(e) is an algebra homomorphism (called augmentation). Consider also the dual morphisms ∆ := µ∗ : A → A ⊗ A

(called comultiplication) and

σ := ι∗ : A → A

(called antipode). In follows using group axioms that these define the structure of a Hopf algebra on k[G]. Conversely, a structure of the Hopf algebra on k[G] defines a structure of an algebraic group on G. An easy corollary of Theorem 3.5.1 now is that the categories of (affine) algebraic groups and affine Hopf algebras are contravariantly equivalent.

Example 8.1.3 (i) k[Ga] = k[T ] with ε(T ) = 0, σ(T ) = −T , and ∆(T ) = T ⊗ 1 + 1 ⊗ T . −1 −1 (ii) k[Gm] = k[T,T ] with ε(T ) = 1, σ(T ) = T , and ∆(T ) = T ⊗T . 8.2 First properties 95

(iii) k[GLn] = k[Ti,j | 1 ≤ i, j ≤ n]det with ε(Tij) = δij, σ(Tij) = i+j (−1) Mj,i/ det (where Mj,i is the determinant of the (j, i)minor), and Pn ∆(Ti,j) = k=1 Tik ⊗ Tkj. A rational representation of G in a ﬁnite dimensional k-vector space V is a homomorphism of algebraic groups ρ : G → GL(V ). The notion of a rational representation is equivalent to that of a rational G-module: V is called a rational G-module if it is a G-module in the usual sense and the corresponding representation is rational. From the point of view of Hopf algebras the notion of a G-module is equivalent to the notion of a comodule over the Hopf algebra k[G] (read about this notion somewhere or better yet invent it yourself!)

8.2 First properties Let G be an algebraic group. We note that only one irreducible component of G can pass through the identity element e. Indeed, if X1,...,Xm are the distinct irreducible components of G containing e. The image of the irreducible variety X1 × · · · × Xm under the product morphism is an irreducible subset X1 ...Xm of G, which again contains e. So X1 ...Xm lies in some Xi. On the other hand each of the components X1,...,Xm clearly lies in X1 ...Xm. This forces m = 1. Denote by G◦ this unique irreducible component of G containing e, and call it the identity component of G.

Proposition 8.2.1 Let G be an algebraic group. (i) G◦ is a normal subgroup of ﬁnite index in G, whose cosets are the connected as well as irreducible components of G. (ii) Each closed subgroup of ﬁnite index in G contain G◦.

Proof (i) We have ι(G◦) is an irreducible component of G containing e, so ι(G◦) = G◦. It also follows from the argument preceding the theorem that G◦G◦ = G◦, so G◦ is a (closed) subgroup of G. For any x ∈ G, xG◦x−1 is also an irreducible component of G containing e, so xG◦x−1 = G◦, i.e. G◦ is normal. Its cosets are translates of G◦, hence must also be irreducible components of G. As there are only finitely many irreducible components, it follows that [G : G◦] < ∞. Since the cosets are disjoint, they are also connected components of G. (ii) If H is a closed subgroup of a finite index in G, then H◦ is a closed subgroup of finite index in G◦, and each of its finitely many left 96 Basic Concepts cosets in G◦ is also closed, and so the union of the cosets distinct from H◦ is closed. Hence H◦ is also open in G◦. Since G◦ is irreducible it is connected, whence H◦ = G◦. The algebraic group is called connected if G◦ = G.

Lemma 8.2.2 Let U and V be dense open subsets of G. Then G = UV .

Proof Let x ∈ G. Then xV −1 and U are dense open subsets. So they have to meet, forcing x ∈ UV .

Lemma 8.2.3 Let H < G be a subgroup of an algebraic group G. Then: (i) H¯ is a subgroup of G. (ii) If H is constructible, then H = H¯ . (iii) If H contains a dense open subset of H¯ , then H = H¯ .

Proof (i) As ι is a homeomorphism, we have ι(H¯ ) = ι(H) = H¯ . Simi- larly, translation by x is a homeomorphism, so xH = xH¯ , i.e. HH¯ ⊂ H¯ . Therefore, if x ∈ H¯ , we have Hx ⊂ H¯ , so Hx¯ = Hx ⊂ H¯ . (ii),(iii) If H is constructible, it contains a dense open subset U of H¯ , see Problem 5.5.3. Then H is also open in H¯ , as H is a union of translates of U. By Lemma 8.2.2, H¯ = HH = H.

Corollary 8.2.4 Let A, B be closed subgroups of G. If B normalizes A, then AB is a closed subgroup of G.

Proof It is clear that AB is a subgroup. Moreover, it the image of A×B under the product morphism, hence constructible by Theorem 5.3.3, hence closed by the lemma.

Lemma 8.2.5 Let ϕ : G → H be a morphism of algebraic groups. Then: (i) ker ϕ is a closed subgroup of G. (ii) im ϕ is a closed subgroup of H. (iii) ϕ(G◦) = ϕ(G)◦. (iv) dim G = dim ker ϕ + dim im ϕ.

Proof (i) follows from the continuity of ϕ and (ii) follows from The- orem 5.3.3 and Lemma 8.2.3(ii). Now, ϕ(G◦) is closed by (ii) and irreducible, hence lies in ϕ(G)◦. Being of ﬁnite index in ϕ(G), it must equal ϕ(G)◦, thanks to Proposition 8.2.1(ii). Finally, Theorem 5.3.1(ii) 8.2 First properties 97 implies that dim G−dim ϕ(G) = dim ϕ−1(x) for ‘most’ points x ∈ ϕ(G). But all ﬁbers ϕ−1(x) are isomorphic to ker ϕ, so we have (iv).

Proposition 8.2.6 Let (Xi, ϕi)i∈I be a family of irreducible varieties and morphisms ϕi : Xi → G such that e ∈ Yi := ϕi(Xi) for all i. Let H be the smallest subgroup of G containing all Yi. Then: (i) H is closed and connected. (ii) H = Y ε1 ...Y εn for some a , . . . , a ∈ I and ε , . . . , ε ∈ {±1}. a1 an 1 n 1 n

−1 Proof We may assume that the sets Yi occur among the Yj. Note that n ¯ for each a := (a1, . . . , an) ∈ I , Ya := Ya1 ...Yan is irreducible, hence Ya is irreducible, too. Obviously YaYb = Y(a,b). ¯ ¯ Moreover, YaYb ⊂ Y(a,b). Indeed, for x ∈ Yb, the homeomorphism (of ¯ ¯ G) y 7→ yx sends Yb to Y(a,b), hence Yb into Y(a,b), i.e YaYb ⊂ Y(a,b). Now ¯ ¯ x ∈ Ya sends Yb into Y(a,b), hence Yb as well. Now choose the tuple a such that dim Ya is maximal. As e ∈ Ya, we ¯ ¯ ¯ have for any b that Ya ⊂ YaYb ⊂ Y(a,b). Equality holds by dimensions, so Y¯b ⊂ Y¯a for every b, and Y¯a is closed under multiplication. Choosing −1 ¯ ¯ b such that Yb = Ya , we also have Ya stable under inversion. So Ya is a group. Since Ya is constructible, it contains a dense open subset of Y¯a, whence Y¯a = YaYa in view of Lemma 8.2.2. Finally, we claim that H = Y¯a. It is clear that H is contained in Y¯a, as we know that each Yb ⊂ Y¯a. Since H ⊃ Ya, we have H¯ = Y¯a. Finally, H ⊃ Ya also implies that H contains a dense open subset of H¯ , so H is closed by Lemma 8.2.3(iii).

Corollary 8.2.7 Assume that (Gi)i∈I is a family of closed connected subgroups of G. Then the group H generated by them is closed and connected. Furthermore, H = Ga1 ...Gan for some a1, . . . , an ∈ I.

Example 8.2.8 It is easy to see that the groups Gm, Ga, GLn are connected. It is less obvious that SLn, Sp2n, and SOn are connected. This can be deduced using Corollary 8.2.7 and some group theory. For example the group SLn is known to be generated by transvections. It follows that the subgroups Gij = {E + tEij | t ∈ k} generate SLn. This transvection subgroups are closed and isomorphic to Ga, hence connected. For Sp2n, let V be the 2n-dimensional vector space on which Sp2n acts, and (·, ·) be the non-degenerate symplectic bilinear form preserved by the group. For v ∈ V \{0} deﬁne the symplectic transvection group Gv to consist of all linear transformations of the form 98 Basic Concepts w 7→ w + t(w, v)v (t ∈ k). It remains to use the known fact that the Gv generate Sp2n. A similar proof is available for SOn. As SOn is of index 2 in On, it follows that it is the identity component of On.

Corollary 8.2.9 Let H and K be closed subgroups of G with H conncted. Then the commutator group (H,K) generated by all commutators [h, k] with h ∈ H, k ∈ K , is closed and connected.

Proof Take the index set I in the proposition to be K and the maps −1 −1 ϕk : H → G to be the maps h 7→ hkh k (k ∈ K).

Example 8.2.10 Recall the deﬁnition of the derived series

G = G(0) ≥ G(1) ≥ ... of a group G: G(0) = G, G(i+1) = (G(i),G(i)). The group G is the called solvable if G(i) = {e} for some i. In case G is a connected algebraic group, each of the derived subgroups are closed connected subgroup of G. So either G(i+1) = G(i) or dim G(i+1) < dim G(i). Thus we see that for algebraic groups the derived series stabilizes after ﬁnitely many steps. Similar remarks apply to nilpotent algebraic groups.

8.3 Actions of Algebraic Groups Let G be an algebraic group and X be a variety (not necessarily aﬃne). We say that G acts on X, or that X is a G-variety, if we are given a morphism G × X → X, (g, x) 7→ gx of varieties that makes X into a G-set in the usual sense. If the G-action on X is transitive, X is called a homogeneous space.

Lemma 8.3.1 Let G act on X. Let Y,Z be subsets of X with Z closed.

(i) The set {g ∈ G | gY ⊂ Z} is closed; in particular NG(Z) := {g ∈ G | gZ ⊂ Z} is closed.

(ii) For each x ∈ X the stabilizer Gx is a closed subgroup of G; in particular, CG(Y ) := {g ∈ G | gy = y for any y ∈ Y } is closed. (iii) The ﬁxed point set Xg of g ∈ G is closed in X; in particular XG is closed. 8.3 Actions of Algebraic Groups 99

Proof (i) For each y ∈ X the orbit map fy : G → X, g 7→ gy is a −1 morphism. So fy (Z) is closed in G. Now note that

−1 {g ∈ G | gY ⊂ Z} = ∩y∈Y fy (Z).

(ii) Observe that Gx = {g ∈ G | g{x} ⊂ {x}} and apply (i). (iii) Consider the morphism ψ : X → X × X, x 7→ (x, gx). Then Xg is the inverse image under ψ of the diagonal, which is closed, since X is a variety.

Remark 8.3.2 The lemma shows that things like centralizers of subsets, normalizers of closed subsets, ﬁxed point sets , etc. are closed. However orbits themselves are not closed in general. In fact the structure of orbits of an algebraic group on a variety can be very interesting. Also, connectedness of centralizers and normalizers is not to be taken for granted.

Theorem 8.3.3 Let G act on X. Then each orbit is smooth, locally closed subset subset of X, whose boundary Gx − Gx is a union of orbits of strictly smaller dimension. In particular, orbits of minimal dimension are closed (so closed orbits exist). If G is connected, the orbits are irreducible.

Proof Let O = Gx. As the image of G under the orbit map, O is constructible, hence contains an open dense subset U of O¯. (Also, O is irreducible if G is connected.) But G acts transitively on O (leaving O¯ stable), so O = ∪g∈GgU is open in O¯, and O is smooth. Therefore O¯ − O is closed and of strictly lower dimension than dim O¯ = dim O. Being G-stable, this boundary is the union of other G-orbits.

n Example 8.3.4 Let G = GLn = GL(V ) where V = k (viewed as an aﬃne n-space). There are just two orbits of G on V : the ponit {0} and the rest V −{0}, an open orbit of dimension n. What can you say about stabilizers in this action? More generally, if V is a rational G-module over an arbitrary algebraic group G, then v 7→ gv deﬁnes a structure of G-variety on V =∼ Adim V .

Example 8.3.5 Again take G = GLn = GL(V ) and define the G-action on P(V ) via ghvi = hgvi (here hvi denote the line spanned by a non-zero vector v ∈ V ). In other words this is just the natural action of GLn on 100 Basic Concepts the lines of V , which is transitive by linear algebra. What can you say about stabilizers in this action? In order to check that this is an action in the sense of algebraic groups, we need to check that the corresponding map ρ : G × P(V ) → P(V ) is a morphism of varieties. For this we employ the Affine Criterion (Theorem 4.2.4) with the usual affine open subsets Vi of P(V ) (0 ≤ i ≤ −1 n n n), and Ui = ϕ (Vi) (Here ϕ : G × P → P is the action map.) Let V 0 = V − {0} be the non-trivial G-orbit from Example 8.3.4. it is easy to check using Affine Criterion that the map V 0 → P(V ), v 7→ hvi is a G-equivariant morphism of varieties.

Example 8.3.6 The natural actions of G = GLn = GL(V ) on the Grassmann variety Gd(V ) and the ﬂag variety F are transitive by linear algebra. These actions are morphic as they are just restrictions of the action of G on P(Λd(V )) and P(Λ1(V )) × · · · × P(Λn−1(V )) × P(Λn(V )), respectively. What can you say about stabilizers in this actions?

Lemma 8.3.7 Let G be a connected algebraic group and X,Y be homogeneous spaces over G. Suppose ϕ : X → Y is a G-equivariant morphism. Set r = dim X − dim Y . Then:

(i) ϕ is surjective and open. (ii) for each closed irreducible subset W ⊂ Y all irreducible components of ϕ−1(W ) have dimension r + dim W .

Proof Surjectivity is clear. Now, it follows from (ii) and Theorem 5.4.1 that ϕ is open. It remains to prove (ii). By Theorem 5.3.1, there is an open set U ⊂ Y such that for each irreducible closed subset W ⊂ Y meeting U, the components of ϕ−1(W ) meeting ϕ−1(U) have dimensions dim W + r. Since G acts transitively on Y and X, the G-translates of U cover Y and the G-translates of ϕ−1(U) cover X. This implies (ii).

8.4 Linear Algebraic Groups

A linear algeraic group is a closed subgroup of some GLn. The following theorem can be thought of as the analogue of the famous theorem that any ﬁnite group is a subgroup of some symmetric group Sn.

Theorem 8.4.1 Every (affine) algebraic group is linear. 8.4 Linear Algebraic Groups 101 To prove the theorem we need to find a finite dimensional vector space on which G acts, and the only place we can look for it is inside the regular module k[G]. Given g ∈ G, the map G → G, h 7→ hg is a morphism of varieties, whose dual map is ρg : k[G] → k[G], where

ρg(f)(h) = f(hg)(f ∈ k[G], h ∈ G). This defines a representation ρ of G in the (usually infinite dimensional space) k[G], called (right) regular representation or representation by right translations of functions. The left regular representation λ is defined similarly via

−1 λg(f)(h) = f(g h)(f ∈ k[G], h ∈ G). The antipode map is actually an isomorphism of the left and right regular representations, so we will usually refer to it as the regular representation and use the right one if we need to write some formulas. The following lemma will help us to deal with the problem of inﬁnite dimensionality of k[G].

Lemma 8.4.2 The regular representation is locally ﬁnite dimensional, i.e. every element of k[G] is contained in a ﬁnite dimensional submodule.

Proof Let us take a non-zero f ∈ k[G]. Let W be the subspace of k[G] spanned by all right translations ρgf. We need to show that W is finite Pn dimensional. Write ∆f = i=1 fi ⊗ gi. Let X be the finite dimensional ubspace of k[G] spanned by all fi. Now consider x ∈ G. We have n X (ρxf)(h) = f(hx) = (∆f)(h, x) = fi(h)gi(x). i=1 Pn Hence ρxf = i=1 gi(x)fi ∈ X. Hence W ⊂ X and W is finite dimensional.

Proof of the theorem Choose linearly independent generators f1, . . . , fn of the algebra k[G]. Applying the lemma, we may assume (adding ﬁnitely many more generators if necessary) that the span E of the fi is invariant under all right translations. Now consider the restriction

ψ : G → GL(E), x 7→ ρx|E of ρ. P Fix i and write ∆fi = j gj ⊗ hj with gj linearly independent and P hj 6= 0. As in the proof of the lemma, ρxfi = hj(x)gj for all x ∈ G, 102 Basic Concepts which implies gj ∈ E, so we can write X ∆fi = fj ⊗ hij (1 ≤ i ≤ n). (8.1) j Then the coordinates of the matrix of ψ(x) with respect to the basis f1, . . . , fn are hij(x). Hence ψ is a morphism of varieties. P Next notice that fi(x) = fi(ex) = j fj(e)hi,j(x), so X fi = fj(e)hi,j. (8.2) j

If ψ(x) = e, then hi,j(x) = δi,j, so fi(x) = fi(e) for all i, whence x = e, as fi’s generate k[G]. By Lemma 8.2.5(ii), G0 := im ψ is a closed subgroup of GL(E). To complete the proof, we need only to show that ψ : G → G0 is an isomorphism of varieties, i.e. ψ∗ : k[G0] → k[G] is an isomorphism of algebras. ∗ As ψ is surjective, ψ is injective. On the other hand, let tij be coordi- 0 ∗ nate functions on GL(E) restricted to G . Note that ψ (tij) = hij, and ∗ the hij generate k[G] in view of (8.2), so ψ is surjective.

8.5 Problems

Problem 8.5.1 Let A be a ﬁnite dimensional k-algebra. Show that Aut(A) is a closed subgroup of GL(A).

Solution. Aut(A) is the stabilizer of an element t ∈ A∗ ⊗ A∗ ⊗ A, see the proof of Corollary 9.5.2.

Problem 8.5.2 Describe Aut(Gm), Aut(Ga), and End (Gm, Gm). ∼ Solution. Working with k[G], we get Aut(Gm) = Z2, where the only non- −1 ∼ trivial automorphism is z 7→ z . Moreover, End (Gm) = Z with m ∈ Z m ∼ × corresponding to the endomorphism z 7→ z . Finally, Aut(Ga) = k , with a ∈ k× corresponding to the endomorphism z 7→ az.

Problem 8.5.3 Closed subset of G containing e and closed under multiplication is a subgroup of G.

Solution. Let X be the subset and x ∈ X. Consider the morphism ϕ : X → X, y 7→ yx. It suﬃces to show that this morphism is surjective, as then e is in the image, and the result follows. 8.5 Problems 103 In order to prove that ϕ is surjective, let Z be an irreducible component of X of maximal dimension. Then ϕ(X) is irreducible of the same dimension, as ϕ is the restriction to X of an autmorphism of G. So ϕ(Z) must be an irreducible component of X. This proves that ϕ permutes irreducible components of X. As X is one-to-one, this argument can now be applied again to the irreducible components of the next largest dimension, etc.

Problem 8.5.4 Let N < GLn be the group of monomial matrices, i.e. matrices having precisely one non-zero entry in each column and each ◦ entry. Prove that N is the subgroup of all diagonal matrices in GLn. Solution. Humphreys, problem 7 after section 7. The group D of diagonal matrices is connected, and [N : D] is ﬁnite.

1 0 Problem 8.5.5 Show that the subgroup of GL ( ) geberated by 2 C 0 −1 1 1 and is not closed. 0 −1 ∼ 1 Solution. Let X = A ⊂ GL2(C) be the closed subset which consists of all upper unitrangular matrices. Note that our subgroup intersects X at the subset of all upper-unitrangular matrices with integer entries in the corner. This is not closed, as Z ⊂ A1 is not closed.

Problem 8.5.6 Let G be a connected algebraic group. Prove that any ﬁnite normal subgroup H lies in the center of G. Solution. If h ∈ H, then the image of the morphism G → G, x 7→ xhx−1 is connected and contained in H, so the image is trivial.

Problem 8.5.7 True or false? Let ϕ : G → H be a morphism of algebraic groups which is an isomorphism of abstract groups. Then ϕ is an isomorphism of algebraic groups.

Solution. False: consider Fr : Gm → Gm or Problem 8.5.8(iii).

Problem 8.5.8 We have A := k[SL2] = k[T11,T12,T21,T22]/(T11T22 − T12T21 − 1) = k[t11, t12, t21, t22](tij denoting the image of Tij). Let B be the subalgebra of A generated by all products tijtkl. (i) Show that B be is a Hopf subalgebra of A and deduce that that there is an algebraic group PSL2 whose algebra is B. Show that 104 Basic Concepts the inclusion map B → A defines a surjective homomorphism of algebraic groups SL2 → PSL2 with kernel of order at most 2. (ii) If char k 6= 2, then B is the algebra of functions f ∈ A such that f(−X) = f(X) for all X ∈ SL2. (iii) If char k = 2, then the homomorphism of (i) defines an isomorphism of underlying abstract groups but is not an isomorphism of algebraic groups. Solution. (i) That B is a Hopf subalgebra is easily checked using explicit formulas for coproduct and antipode. Also B is a reduced finitely generated k-algebra, so it corresponds to an algebraic group by general principles. Also, the inclusion map ι : B → A, being a Hopf algebra map, defines a surjective homomorphism ι∗ : SL2 → PSL2. Now B is generated by the elements

2 2 2 2 t11, t11t12, t11t21, t11t22, t12, t12t22, t21, t21t22, t22, as t11t22 = t12t21 + 1. Now, using the counit, we see that the identity e 2 2 in PSL2 is deﬁned by equations t11(e) = 1, t22(e) = 1, t11t22(e) = 1 and tijtkl(e) = 0 for all other generators. So A = (aij) maps to e under ι∗ if 2 2 and only if a11 = 1, a22 = 1, a11a22 = 1 and aijakl = 0 for all other pairs of indices corresponding to the generators. It follows that the kernel of ι∗ is ±I. (ii) Direct check. (iii) If char k = 2, ι∗ is bijective. Of course it is not an isomorphism since ι is not surjective.

Problem 8.5.9 Let X be a G-variety and a : G × X → X is the action map. Define the left action of G on k[X] via (gf)(x) = f(g−1x)(g ∈ G, x ∈ X, f ∈ k[X]). Note that this yields a representation of abstract group G in k[X]. (i) The representation is locally finite dimensional. (ii) A finite dimensional subspace V ⊂ k[X] is G-stable if and only if a∗(V ) ⊂ k[G] ⊗ V . If so, the action of G on V defines a rational representation of G. (iii) There is a sequence of finite dimensional G-submodules Vi ⊂ k[G] such that V1 ⊂ V2 ⊂ ... and k[X] = ∪iVi. ∗ P Solution. Take f ∈ k[X]. If a : k[X] → k[G]⊗k[X] maps f to i hi⊗fi, P −1 then gf = i hi(g )fi, which implies (i) and (ii). Now (ii) is a general fact on countably dimensional locally finite modules. 9 Lie algebra of an algebraic group

9.1 Definitions Let G be an algebraic group and A = k[G]. We will consider the Lie algebra Der(A) of k-derivations A → A with respect to the bracket [δ1, δ2] = δ1 ◦ δ2 − δ2 ◦ δ1. A derivation δ ∈ Der(A) is called left-invariant if it commutes with left translations, i.e. δ ◦ λx = λx ◦ δ for all x ∈ G. The left invariant derivations of A form a Lie subalgebra of Der(A), called the Lie algebra of G and denoted L(G). (Using right invariant derivations here would lead to an isomorphic object). Let us denote by g the tangent space TeG. We claim that g can be naturally identified with L(G) as vector spaces. Recall that TeG can be defined as the the derivations of A at e. Define a k-linear map θ : L(G) → g by

(θδ)(f) = (δf)(e)(δ ∈ L(G), f ∈ A).

We claim that θ is an isomorphism of vector spaces. In order to prove this we construct the inverse map η : g → L(g) sending a tangent vector X to a derivation ∗X called right convolution by X and deﬁned by

(f ∗ X)(x) = X(λx−1 f)(x ∈ G, f ∈ A). It is a straightforward check that ∗X is indeed a left invariant derivation of A and that η is k-linear. Finally, η is inverse to θ:

(f ∗ θ(δ))(x) = θ(δ)(λx−1 f) = δ(λx−1 f)(e) = λx−1 (δf)(e) = (δf)(x),

θ(∗X)(f) = (f ∗ X)(e) = X(λe−1 f) = X(f) (for X ∈ g, δ ∈ L(G), f ∈ A, x ∈ G). From now on we are going to identify L(G) with g via the isomorphisms θ and η. For example, g is a Lie algebra with respect to the

105 106 Lie algebra of an algebraic group bracket deﬁned as follows:

[X,Y ](f) = ((f ∗ Y ) ∗ X − (f ∗ X) ∗ Y )(e) = X(f ∗ Y ) − Y (f ∗ X).

We give another deﬁnition of [X,Y ] in terms of the coproduct ∆. Deﬁne

X · Y : A → k, f 7→ (X ⊗ Y ) ◦ ∆(f).

P 0 If ∆(f) = i fi ⊗ fi , then

X 0 f ∗ X = fiX(fi ), i whence it is easy to see that (X · Y )(f) = ((f ∗ Y ) ∗ X))(e). So

[X,Y ] = X · Y − Y · X.

This deﬁnition of the bracket makes the following easy to check:

Theorem 9.1.1 If ϕ : G → G0 is a homomorphism of algebraic groups, then dϕ : g → g0 is a homomorphism of Lie algebras.

If H is a closed subgroup of an algebraic group G, the inclusion η : H → G is an isomorphism onto a closed subgroup, with η∗ : k[G] → k[H] = k[G]/I being the natural projection. Therefore, dη identifies h with the Lie subalgebra of g consisting of those X ∈ g for which X(I) = 0. We will always identify h with a Lie subalgebra of g in this way. Now, let ϕ : G → G0 be a morphism of algebraic groups, H0 < G0 is a closed subgroup, and ϕ(H) ⊂ H0. Then ϕ|H can be considered as a morphism H → H0, so its differential d(ϕ|H) is a Lie algebra homomorphism h → h0. It follows from the definitions that

(dϕ)|h : h → h0 = d(ϕ|H). (9.1)

Lemma 9.1.2 Let H be a closed subgroup of an algebraic group G and I = I(H) / k[G]. Then h = {X ∈ g | I ∗ X ⊂ I}.

Proof If f ∈ I, X ∈ h, and x ∈ H, then (f ∗ X)(x) = X(λx−1 f) = 0 since λx−1 f ∈ I. Conversely, if I ∗ X ⊂ I and f ∈ I, then (f ∗ X)(e) = X(λe−1 f) = X(f) = 0, forcing X ∈ h.

Lemma 9.1.3 Let ρ : G → GL(V ) be a rational representation and dρ : g → gl(V ) be the corresponding Lie algebra representation. If W ⊂ V is a G-invariant subspace then W is also g-invariant. 9.2 Examples 107 Proof If we extend a basis of W to a basis of V , then the matrix of ∗ ∗ any ρ(x) has the form , and so the matrix of any dρ(X) has the 0 ∗ same form.

9.2 Examples

Example 9.2.1 If G = Ga, then g is 1-dimensional, so its bracket is trivial.

Example 9.2.2 Let G = GLn. Its tangent space at e has as a basis ∂ the set of partial derivatives |e (evaluated at e). The coordinates xij ∂Tij with repsect to this basis can be arranged in a square matrix. So we can think of tangent vectors X as square matrices (xij), where xij = X(Tij). P With this convention (X · Y )(Tij) = l xilylj. In other words, X · Y is the usual matrix product XY . Thus g = gln(k).

Example 9.2.3 The Lie algebra sln(k) of SLn < GLn consists of all P ∂ matrices in gl (k) of trace 0. Indeed, let X = (aij) = aij |e be n ∂Tij a tangent vector. Then X ∈ sln(k) if and only if X(det) = 0, which is equivalent to tr X = 0.

t 2 Example 9.2.4 The group Sp2n < GL2n is Z(x Jx−x) (4n polynomial equations written as one matrix equation). So the Lie algebra sp2n(k) t consists of all matrices X ∈ gl2n(k) with X(x Jx − x) = 0. This is t equivalent to X J + JX = 0 (compute!). Compute dim sp2n(k).

t 2 Example 9.2.5 The group On < GL2n is Z(xx − 1) (n polynomial equations written as one matrix equation). So the Lie algebra son(k) t consists of all matrices X ∈ gln(k) with X + X = 0.

Example 9.2.6 The Lie algebra u of the subgroup Un < GLn of upper unitriangular matrices consists of all strictly upper triangular matrices in gln(k).

Lemma 9.2.7 Let G be an algebraic group with product µ : G × G → G and inverse ι : G → G. Then for all X,Y ∈ g:

(i) dµ(e,e)(X,Y ) = X + Y ; (ii) dιe(X) = −X; 108 Lie algebra of an algebraic group

Proof Let (X,Y ) ∈ g ⊕ g = T(e,e)G × G, and Z := dµ(e,e)(X,Y ). If f ∈ P 0 P 0 0 k[G] and ∆(f) = i fi ⊗ fi , then Z(f) = i(X(fi)fi (e) + fi(e)Y (fi )), cf. the proof of Proposition 6.1.5. On the other hand, we have

X 0 X 0 f = fi(e)fi = fi (e)fi. i i (you should have checked that when you checked the axioms of Hopf algebra for k[G], but it’s not too late now). So Z(f) = (X + Y )(f), giving (i). Consider the composite G → G × G → G, g 7→ (g, ι(g)) 7→ gι(g) = e. The composite is a constant function, so its differential is zero. But the differential of a composite is the composite of the differentials, so applying (i), we have 0 = d ide +dιe = id +dιe, whence (ii).

Lemma 9.2.8 Let E ⊂ k[G] be a ﬁnite dimensional subrepresentation of the (right) regular representation ρ of G, and ψ : G → GL(E) be the restriction of ρ to E. Then dψ(X)(f) = f ∗ X for f ∈ E.

P Proof Pick a basis {f1, . . . , fn} of E. Let ∆(fi) = j fj ⊗ mij, see P (8.1). Then ρx(fi) = j mij(x)fj. So the matrix of ψ(x) in our basis is (mij(x)). Note, moreover, that X λx−1 fi = fj(x)mij. (9.2) j

Now, let X ∈ g. By deﬁnition, the (i, j)entry of the matrix dψ(X) is ∗ X(ψ (Tij)) = X(mij). On the other hand, using (9.2), we get X (fi ∗ X)(x) = X(λx−1 fi) = fj(x)X(mij), j which completes the proof.

9.3 Ad and ad −1 Fix x ∈ G. Let Int x : G → G, y 7→ xyx . The diﬀerential d(Int x)e is a Lie algebra automorphism denoted

Ad x : g → g.

The image of Ad is a (closed connected) subgroup of GL(g)) denoted Ad G. 9.3 Ad and ad 109

−1 Example 9.3.1 Let G = GLn. Then Ad x(X) = xXx (for X ∈ g = gln(k)). Hence for any closed subgroup H < G, its Lie algebra h, and x ∈ H, Ad x : h → h is conjugation by x too. ∗ For the proof, let us compute (Int x) (Tij):

∗ −1 X −1 (Int x) (Tij)(g) = Tij(xgx ) = xikTkl(g)(x )lj. k,l

Hence

∗ X −1 (Int x) (Tij) = xik(x )ljTkl. k,l

Now, the ij-entry of Ad x(X) is

∗ X −1 Ad x(X)(Tij) = X((Int x) (Tij)) = xik(x )ljX(Tkl), k,l which is the ij-entry of xXx−1.

Theorem 9.3.2 Ad is a rational representation of G in (the vector space) g (called the adjoint representation of G).

Proof Embed G as a closed subgroup of some GLn. Then by Exam- ple 9.3.1, Ad x is a conjugation by x, which implies that Ad : G → GL(g) is a morphism of varieties.

Let ad : g → gl(g) be the adjoint representation of Lie algebra, i.e.

ad X(Y ) = [X,Y ](X,Y ∈ g).

Theorem 9.3.3 The diﬀerential of Ad is ad.

Proof Using embedding of G into some GLn and (9.1), it suﬃces to check the result for G = GLn. Note that Ad x is the image of x under the map

(1,ι) µ G −→ G × G −→σ×τ GL(g) × GL(g) −→ GL(g), where σ(x) (resp. τ(x)) is the left (resp. right) multiplication by x in g. Since the entries of σ(x) and τ(x) are linear polynomials in the entries of x, it follows that dσ(X) (resp. dτ(X)) is a left (resp. right) multiplication by X. Now the result follows from Lemma 9.2.7. 110 Lie algebra of an algebraic group 9.4 Properties of subgroups and subalgebras

Lemma 9.4.1 If H is a closed normal subgroup of an algebraic group G, then h is an ideal g.

Proof We have Int x stabilizes H for all x ∈ G. Hence Ad x stabilizes h for all x ∈ G. If we extend a basis of h to a basis of g, then the matrix ∗ ∗ of Ad x therefore has the form (x ∈ G), and so the matrix of 0 ∗ d(Ad)(X) = ad X has the same form (X ∈ g).

Lemma 9.4.2 If H is a closed subgroup of an algebraic group G and N = NG(H), then n ⊂ ng(h).

Proof Note that N is closed in view of Lemma 8.3.1(i). Applying Lemma 9.4.1 to the normal subgroup H of N, we see that h is an ideal of n, i.e. n normalizes h. For x ∈ G denote

−1 −1 γx : G → G, y 7→ yxy x .

Lemma 9.4.3 (dγx)e(X) = X − Ad x(X).

Proof Consider ﬁrst the morphism ψ : G → G, y 7→ xy−1x−1. As ψ = Int x ◦ ι, we have

dψe(X) = d(Int x) ◦ dιe(X) = Ad x(−X) = − Ad(X).

Now γx can be realizete as the composite

(1,ψ) µ G −→ G × G −→ G.

So (dγx)e(X) = dµ(e,e)(X, dψe(X)) = X − Ad x(X).

Lemma 9.4.4 Let x ∈ G. Then L(CG(x)) ⊂ cg(x) := {X ∈ g | Ad x(X) = X}. If G = GLn, then equality holds.

−1 Proof Note that the Lie algebra L(CG(x)) of the ﬁber γx (e) = CG(x) maps to zero under the map (dγx)e. Now use Lemma 9.4.3. In case of GLn the ﬁxed points of Ad x in g are just the matrices commuting with x, so CG(x) is a principal open set in cg(x), containing e, which implies the result. 9.5 Automorphisms and derivations 111 Lemma 9.4.5 Let ρ : G → GL(V ) be a rational representation, and dρ : g → gl(V ) be the corresponding representation of the Lie algebra. If v ∈ V , let CG(v) = {x ∈ G | xv = v} and cg(v) := {X ∈ g | Xv = 0}. Then L(CG(v)) ⊂ cg(v).

Proof Note that x 7→ xv is a morphism G → V constant on CG(v), so dρe is zero on the Lie algebra L(CG(v)).

Lemma 9.4.6 Let A and B be closed subgroups of G, and let C be the closure of the subgroup C = (A, B) generated by the commutators. The its Lie algebra c contains all elements of the form [X,Y ], Y − Ad x(Y ), X − Ad y(X)(x ∈ A, X ∈ a, y ∈ B,Y ∈ b). In particular, if H is the closure of (G, G), then h ⊃ [g, g].

Proof For x ∈ A, γx maps A to C, so the diﬀerential 1 − Ad x maps b to c. This yields all elements of the second type listed, and similarly for the third type. Next for X ∈ a consider the morphism ϕ : B → c deﬁned by ϕ(y) = X −Ad y(X). Since ϕ maps e to 0, we have dϕe(Y ) = − ad Y (X) = −[Y,X] = [X,Y ].

Remark 9.4.7 Inclusions in Lemmas 9.4.2, 9.4.4, 9.4.5, and 9.4.6 can be proper in positive characteristic and are equalities in characteristic 0.

9.5 Automorphisms and derivations

Lemma 9.5.1 Let V and W be rational G-modules. Then (i) g acts on V ∗ by the rule Xf(v) = −f(Xv) for f ∈ V ∗, v ∈ V,X ∈ g. (ii) g acts on V ⊗ W by the rule X(v ⊗ w) = (Xv) ⊗ W + v ⊗ (Xw) for v ∈ V, w ∈ W, X ∈ g.

Proof (i) We fix a basis of V and write the action of x ∈ G as a matrix. Then the matrix of x acting on the dual basis of V ∗ is the transpose inverse matrix. We know that the differential of x 7→ x−1 is X 7→ −X, t t while the map x 7→ x of GLn has the differential X 7→ X on gln. This implies the result.

(ii) Fix bases {v1, . . . , vn} of V and {w1, . . . , wm} of W , and let ρ1 : G → GLn, ρ2 : G → GLm be the corresponding matrix representations. 112 Lie algebra of an algebraic group

If ρ1(x) = (aij) and ρ2(x) = (brs), then the matrix (ρ1 ⊗ ρ2)(x) has entry airbjs in the (i, j)row and (r, s)column. So the representation G → GLmn factors as the composite of two morphisms

(ρ1,ρ2) G −→ GLn × GLm → GLmn, where the second map is given in coordinates via Zij,rs = XirYjs. It is easy to compute the diﬀerential (at (e, e)) of the second morphism—it maps a pair of matrices ((cij), (drs)) ∈ gln ⊕ glm to the matrix whose entry in row (i, j) and column (r, s) is δjscir + δirdjs. This implies the rule asserted in (ii).

Corollary 9.5.2 Let B be a ﬁnite dimensional k-algebra (not necessarily associative), and let G be a closed subgroup of GL(B), consisting of algebra automorphisms. Then g consists of derivations of B.

∗ ∗ Proof Let t ∈ B ⊗ B ⊗ B = Homk(B ⊗ B, B) be the multiplication on B. Note that x ∈ GL(B) is an automorphism of B if and only if t is an invariant of x. So t is an invariant of G, whence it is an invariant of g, see Lemma 9.4.5. This is equivalent to the fact that g consists of derivations of B.

9.6 Problems

Problem 9.6.1 Let H be a closed subgroup of G = GL(V ), h ⊂ gl(V ) be its Lie algebra, v ∈ V , and W ⊂ V be a vector subspace. (i) If H leaves W stable, then so does h. Is the converse true? (ii) If H leaves v stable, then h kills v. Is the converse true?

(iii) Set GW := {x ∈ G | x(W ) ⊂ W }, gW := {X ∈ g | X(W ) ⊂ W }. Then L(GW ) = gW .(Hint: L(GW ) ⊂ gW by (i). Now, use explicit descriptions of GW and gW using matrices and dimensions). (iv) Set Gv := {x ∈ G | xv = v}, gv := {X ∈ g | Xv = 0}. Then L(Gv) = gv.

Problem 9.6.2 Prove that Z(G) ⊂ ker Ad.

Problem 9.6.3 Let char k = p > 0 and G ⊂ GL3 consist of all matrices a 0 0 of the form 0 ap b with a 6= 0. Observe that g consists of all 0 0 1 9.6 Problems 113 a 0 0 matrices 0 0 b and is commutative. Moreover, {e} = Z(G) ( 0 0 0 ker Ad ( G.

Problem 9.6.4 Let char k = 2, G = SL2, and B the group of all upper triangular matrices in G. Then NG(B) = B, whereas ng(b) = g.

Problem 9.6.5 Let H < G be a closed subgroup and x ∈ G. Then Ad x(h) = L(Int x(H)).

Problem 9.6.6 Deﬁne P GLn := Ad GLn and PSLn := Ad SLn. The centers of GLn and SLn consist of all scalar matrices contained in ∼ ∼ these groups. As abstract groups P GLn = GLn/Z(GLn) and PSLn = SLn/Z(SLn). If the characteristic p of k divides n, then Z(SLn) = {1}, but SLn is not isomorphic to PSLn as algebraic groups!!!!

Problem 9.6.7 If char k = p > 0 and X is a left invariant derivation of k[G], then Xp is also a left invariant derivation of k[G]. This gives an extra operation on g, called pth power operation, which makes g into a restricted Lie algebra. (One needs to check a number of axioms here, but never mind...) Compute the pth power operation for G = Ga and G = GLn. 10 Quotients

The main problem addressed in this chapter is as follows: given an algebraic group G and a closed subgroup H, how to endow the quotient G/H with a ‘reasonable’ structure of algebraic variety?

10.1 Construction We start with a linear algebra lemma.

Lemma 10.1.1 Let M be a d-dimensional subspace of a vector space W , x ∈ GL(W ), X ∈ gl(W ). Then L := ΛdM can be considered as a line in ΛdW . (i) xL = L if and only if xM = M. (ii) XL ⊂ L if and only if XM ⊂ M.

Proof Exercise or read it in Humpreys.

Theorem 10.1.2 (Chevalley) Let G be an algebraic group, H < G a closed subgroup. Then there is a rational representation ϕ : G → GL(V ) and a 1-dimensional subspace L of V such that H = {x ∈ G | ϕ(x)L = L} and h = {X ∈ g | dϕ(X)L ⊂ L}.

Proof Let I = I(H) / k[G]. Let W ⊂ k[G] be a ﬁnite dimensional subspace invariant with respect to all ρx and containing a (ﬁnite) generating set for I, see Lemma 8.4.2. Let M = W ∩ I (so M generates I). Note that H = {x ∈ G | ρxI = I}, so M is stable under all ρy for y ∈ H. It follows from Lemmas 9.2.8 and 9.1.3 that M is stable under all ∗Y for Y ∈ h.

114 10.2 Quotients 115

We claim that H = {x ∈ G | ρxM = M} and h = {X ∈ g | M ∗ X ⊂ M}. Indeed, if ρxM = M, then we have

ρxI = ρx(MA) = ρx(M)ρx(A) = MA = I, forcing x ∈ H. If M ∗ X ⊂ M then the product rule implies I ∗ X = (MA) ∗ X ⊂ (M ∗ X)A + M(A ∗ X) ⊂ MA = I, forcing X ∈ h by Lemma 9.1.2. Finally, pass to ΛdW where d = dim M, take ϕ to be the dth exterior power of the representation constructed above, and use Lemma 10.1.1.

Corollary 10.1.3 Let H be a closed subgroup of a connected algebraic group G. Then there exists a quasi-projective variety X that G acts transitively on and a point x ∈ X such that

(i) Gx = H; (ii) the orbit map ψ : G → X, g 7→ gx is separable; (iii) the ﬁbers of ψ are the cosets gH of H in G.

Proof Let V and L = hvi ⊂ V be as in the theorem. Take X to be the G-orbit Ghvi in P(V ) and x = hvi. This is open in its closure, hence it is a quasi-projective variety. By the theorem, H = Gx, and now (iii) is also clear. Finally note that the tangent space to P(V ) at x can be canonically identified with V/hvi, and the tangent space to X at x is a subspace of V/hvi. The differential dψe maps Y ∈ g to Y v + hvi. Now, by the theorem, the kernel of the differential is h. So

dim ker dψe = dim h = dim H = dim G − dim X.

Hence dψe is onto by dimension, and ψ is separable in view of Theo- rem 6.7.1.

10.2 Quotients In this section we will assume that G is a connected algebraic group and H < G a closed subgroup. (The assumption that G is connected is not essential, but we do not want to deal with necessary modifications needed in the non-connected case). A Chevalley quotient of G by H is a variety X together with a surjective separable morphism π : G → X such that the fibers of π are exactly 116 Quotients cosets of H in G. By Corollary 10.1.3 Chevalley quotients exist, but it is not clear if they are unique up to isomorphism. A categorical quotient of G by H is a variety X together with a morphism π : G → X that is constant on all cosets of H in G with the following universal property: given any other variety Y and a morphism ϕ : G → X that is constant on all cosets of H in G there is a unique morphismϕ ¯ : X → Y such that ϕ =ϕ ¯ ◦ π. Now, it is clear that categorical quotients are unique up to unique isomorphism, but it is not clear if they exist. Our goal is to prove that Chevalley quotients are categorical quotients. This will prove that categorical quotients exist and that Cheval- ley quotients are unique. So we need to take a Chevalley quotient (X, π) and check that it has the right universal property. Given a morphism ϕ : G → Y constant on cosets, there is a unique map of sets X → Y such that ϕ =ϕ ¯ ◦ π, since fibers of π are exactly the cosets. But it is very difficult to prove from this point of view that ϕ is a morphism of varieties. So we proceed rather differently.

Theorem 10.2.1 Chevalley quotients are categorical quotients.

Proof Step 1. Let us try to construct a categorical quotient not in the category of varieties but in the more general category of geometric spaces. Define G/H to be the set of cosets of H in G. Let π : G → G/H be the map x 7→ xH. Give G/H the structure of topological space by declaring U ⊂ G/H to be open if and only if π−1(U) is open. Next define a sheaf O of functions on G/H: if U ⊂ G/H is open, let O(U) −1 consist of all functions f on U such that f ◦ π ∈ OG(π (U)). (Check the sheaf axioms!) In order to check the universal property, let ψ : G → Y be a morphism of geometric spaces constant on the cosets of H in G. We get the induced map of sets ψ¯ : G/H → Y, xH 7→ ψ(x). We claim that ψ is a morphism of geometric spaces. For continuity, take an open subset V ⊂ Y , and note that U := ψ¯−1(V ) is open in G/H, as π−1(ψ¯−1(V )) = ψ−1(V ) is ¯∗ open in G. Finally, take f ∈ OY (V ) and show that ψ (f) ∈ OG/H (U). ∗ ¯∗ −1 By definition, we just need to check that π (ψ (f)) ∈ OG(ψ (V )). ∗ ¯∗ ∗ −1 But π (ψ (f)) = ψ f ∈ OG(ψ (V )), as ψ is a morphism of geometric spaces. Step 2. Now, let (G/H, π) be as in step 1, and let (X, ψ) be a Cheval- ley quotient. Using the universal property established above, we get a unique G-equivariant morphism ψ¯ : G/H → X such that ψ = ψ¯ ◦ π, 10.2 Quotients 117 i.e. ψ¯(xH) = ψ(x). We will prove that ψ¯ is an isomorphism of geometric spaces, which will imply that G/H is a variety and that X is a categorical quotient. First of all, it is clear that ψ¯ is bijective. Moreover, by Lemma 8.3.7, the map ψ is open (and continuous), which implies that ψ¯ is a homeomorphism. In order to finish the proof, take an open subset U ⊂ X, −1 a function f ∈ OG(ψ (U)) constant on the cosets, and prove that ∗ f = ψ (g) for some g ∈ OX (U). For simplicity we consider the case U = X when ψ−1(U) = G. The argument for the general case is similar. We show first that there exists a rational function g with the required property, i.e. f = ψ∗(g) in k(G). Consider the morphisms

ϕ 1 π1 G −→ X × A −→ X, where ϕ = (ψ, f). The composite is just ψ. If Y is the closure in X ×A1 of ϕ(G), then Y is irreducible, and π1 induces a surjective morphism η : Y → X. Since ψ is separable, so is η (use ψ∗ = ϕ∗ ◦ η∗). Now, ϕ(G) contains a dense open subset of Y , see Problem 5.5.3. Since f is constant on fibers of ψ, the restriction of η to this open set is injective, as well as dominant and separable. By Theorem 5.4.3, η∗ 1 1 maps k(X) isomorphically onto k(Y ). But π2 : X × A → A induces on Y a morphism g : Y → A1, i.e. a regular function, in particular a rational function. So there exists h ∈ k(X) for which g = η∗h. Finally, notice that ϕ∗g = ϕ∗η∗h = ψ∗h agrees everywhere on G with f. So f = ψ∗h, as desired. Next we want to show that the rational function h ∈ k(X) just constructed is actually a regular function on Y . Since all points of X are simple, Theorem 6.3.2 shows that unless h is everywhere defined on X, 1/h is defined and is equal to 0 at some point. But then ψ∗(1/h) = 1/f must also take the value zero, which is absurd since f ∈ k[G].

We will denote by G/H the categorical quotient of G by the closed subgroup H. We now know that the categorical quotient exists and is unique up to a unique isomorphism. We also know that G/H is a quasi-projective variety and π : G → G/H is separable. Also note that ∼ TeH (G/H) = g/h.

Indeed, the separability of π implies that dπeH is surjective, and it contains h in its kernel. Now use dimensions. 118 Quotients

Example 10.2.2 Let G = GL(V ), {v1, . . . , vn} be a basis of G, and

G = Ghv1i. Then X = Ghv1i = P(V ) is a Chevalley quotient of G by H, see the proof of Corollary 10.1.3. So G/H =∼ P(V ). Similarly, let ∼ ∼ P = Ghv1,...,vdi. Then G/P = Gd(V ). Finally, G/B = F, where B is the stabilizer of a standard ﬂag. In the examples above quotients are projective varieties. Let K := ∼ n Gv1 . Then G/K = A \{0}. This is neither aﬃne nor projective (unless n = 1).

t Example 10.2.3 Let G = GLn and H = On = {g ∈ GLn | g g = I} (assuming char k 6= 2). Let S be the set of all n × n symmetric matrices, and affine variety of dimension n(n + 1)/2. Let S× be the invertible matrices in S, a principal open subset of S, hence also affine of dimension n(n + 1)/2. t Let G act on S by g · x = g xg. Then GI = On, and the action is transitive, as by linear algebra all non-degenerate symmetric bilinear ∼ × forms are equivalent. To prove that GLn/On = S , we just need to prove that the orbit map G → S×, g 7→ gtg is separable. Its differential is the map X 7→ Xt +X. The tangent space to S× at I can be identified with S. Clearly any symmetric matrix can be written in the form Xt +X (characteristic is not 2!). ∼ × Thus GLn/On = S , which is an affine variety.

10.3 Normal subgroups Let G be an algebraic group. A character of G is a homomorphism χ : G → Gm of algebraic groups. We write X(G) for the set of all characters of G. It has a natural structure of an abelian group:

(χ + ψ)(g) = χ(g)ψ(g).

Let V be a rational G-module. For χ ∈ X(G), let

Vχ := {v ∈ V | gv = χ(g)v for all g ∈ G}. P It is easy to see that χ∈X(G) Vχ = ⊕χ∈X(G)Vχ, see Problem 10.4.5. On P the other hand, it is usually not true that V = χ∈X(G) Vχ. But there is ∼ n one important case when it is the case. This is when G = Dn = (Gm) , the group of all diagonal matrices in GLn. This will be established later. Now, let N be a closed normal subgroup of G, and V be a rational G-module. If χ ∈ X(N), then for any g ∈ G we have gVχ ⊂ Vχ0 for 10.3 Normal subgroups 119

0 −1 χ = gχ ∈ X(N). Here gχ(h) := χ(g hg). Indeed, let v ∈ Vχ and h ∈ N. Then hgv = gg−1hgv = χ(g−1hg)gv.

Theorem 10.3.1 Let G be an algebraic group and N ⊂ G be a closed normal subgroup. Then the variety G/N is aﬃne, and G/N × G/N → −1 G/N, (g1N, g2N) 7→ g1g2N, G/N → G/N, gN 7→ g N are morphisms of varieties.

Proof Let us show ﬁrst that (g1N, g2N) 7→ g1g2N is a morphism. The map G×G → G/N, (g1, g2) 7→ g1g2N is a morphism that is constant on cosets of N ×N. Hence by the universal property of the quotients, we get induced a unique morphism G/N × G/N =∼ (G × G)/(N × N) → G/N, see Problem 10.4.1. The proof that gN 7→ g−1N is a morphism is similar (but easier). By Chevalley’s theorem, we can ﬁnd a rational representation ρ : G →

GL(V ) and v ∈ V such that H = Ghvi, and h is the stabilizer of hvi in 0 0 0 g. Let V = ⊕χ∈X(N)Vχ. Note that v ∈ V and V is G-invariant, so we may assume that V = V 0.

Now, let W = {f ∈ End (V ) | f(Vχ) ⊂ Vχ for all χ ∈ X(H)}. Deﬁne a morphism of algebraic groups ψ : G → GL(W ), where

ψ(g)f = ρ(g)fρ(g)−1 (g ∈ G, f ∈ W ).

Let us compute the kernel of ψ: if ψ(g) = id, then ρ(g) stabilizes each Vχ and commutes with End (Vχ), hence by Schur’s lemma ρ(g) acts as scalars on each Vχ. Hence g stabilizes hvi, so g ∈ H. Conversely, if H acts as a scalar on each Vχ, then H ⊂ ker ψ. Note that the image of ψ is a closed—hence aﬃne—subgroup of GL(W ). To show that this is a Chevalley quotient we just need to prove that ψ is separable. For this we show that dψ is onto, or equivalently by dimensions that ker dψ ⊂ h. Let X ∈ ker dψ. Then dψ(X)(f) = dρ(X)f − fdρ(X) = 0, so dρ(X) commutes with all f ∈ W . This implies that dρ(X) acts as a scalar on all Vχ’s, in particular, it stabilizes hvi, hence X ∈ h.

Corollary 10.3.2 Suppose that ϕ : G → H be a separable surjective morphism of algebraic groups and N = ker ϕ. Then ϕ induces an isomorphism G/N =∼ H.

Note in characteristic 0 the separability is automatic. On the other hand, let G = H = GLn, and let ϕ be the Frobenius homomorphism 120 Quotients given by raising matrix entries to the pth power. This is a morphism of algebraic groups and an isomorphism of abstract groups, but the diﬀerential dϕ is the zero map. So ϕ is deﬁnitely not an isomorphism of algebraic groups.

10.4 Problems

Problem 10.4.1 Let H1 < G1,H2 < G2 be closed subgroups of con- ∼ nected algebraic groups. Prove that (G1 × G2)/(H1 × H2) = G1/H1 × G2/H2.

Problem 10.4.2 Prove that GL2n/Sp2n is isomorphic to the aﬃne variety of all invertible 2n × 2n-skew symmetric matrices. (In characteristic 2 a skew symmetric matrix means a symmetric matrix with zeros on the main diagonal).

∼ Problem 10.4.3 Prove that X(SLn) = {0}, X(Ga) = {0}, X(Gm) = Z, X(GLn) = Z.

Problem 10.4.4 Prove that X(G × H) =∼ X(G) ⊕ X(H).

P Problem 10.4.5 Prove that χ∈X(G) Vχ = ⊕χ∈X(G)Vχ

Problem 10.4.6 Let A, B ⊂ G be closed subgroups. Prove that a∩b = L(A ∩ B) if and only if the restriction to A of the canonical morphism π : G → G/B is again separable. (Hint: consult Theorem 12.1.1.)

Problem 10.4.7 Let H be a closed subgroup of a connected algebraic group G. Then H acts naturally on k(G) as a group of automorphisms, and k(G/H) =∼ k(G)H .

Problem 10.4.8 Compute the dimension of the ﬂag variety. 11 Semisimple and unipotent elements

11.1 Jordan-Chevalley decomposition The following result about the additive Jordan decomposition is well known:

Lemma 11.1.1 Let V be a ﬁnite dimensional k-vector space and X ∈ End V .

(i) There exist unique Xs,Xn ∈ End (V ) satisfying the conditions X = Xs + Xn, Xs is semisimple, Xn is nilpotent, and XsXn = XnXs. (ii) There exist polynomials p(T ), q(T ) without constant term such that Xs = p(X),Xn = q(X). In particular Xs,Xn commute with any endomorphism of V which commutes with X.

(iii) If A ⊂ B ⊂ V are subspaces and X maps B to A, then so do Xs and Xn.

(iv) If XY = YX for Y ∈ End V then (X + Y )s = Xs + Ys and (X + Y )n = Xn + Yn. (v) If ϕ : V → W is a linear map and Y ∈ End W such that Y ◦ ϕ = ϕ ◦ X, then Ys ◦ ϕ = ϕ ◦ Xs and Yn ◦ ϕ = ϕ ◦ Xn.

An element x ∈ End V is called unipotent if it is the sum of idV and a nilpotent element, or, equivalently, if the only eigenvalue of x is 1. In characteristic p an element x ∈ End V is unipotent if and only if N xp = 0 for some N. The additive Jordan decomposition implies the multiplicative Jordan decomposition:

Lemma 11.1.2 Let V be a ﬁnite dimensional k-vector space and x ∈ GL(V ).

121 122 Semisimple and unipotent elements

(i) There exist unique xs, xu ∈ End (V ) satisfying the conditions x = xsxu, xs is semisimple, xu is unipotent, and xsxu = xuxs.

(ii) xs, xu commute with any endomorphism of V which commutes with x.

(iii) If A ⊂ V is a subspaces stable under x, then A is stable under xs and xu.

(iv) If xy = yx for y ∈ GL(V ) then (xy)s = xsys and (xy)u = xuyu. (v) If ϕ : V → W is a linear map and y ∈ End W such that y ◦ ϕ = ϕ ◦ x, then ys ◦ ϕ = ϕ ◦ xs and yu ◦ ϕ = ϕ ◦ xu.

We leave the following as an exercise:

Lemma 11.1.3 Let x = xsxu and y = ysyu be Jordan decompositions of x ∈ GL(V ) and y ∈ GL(W ). Then x⊕y = (xs ⊕ys)(xu ⊕yu) and x⊗y = (xs ⊗ ys)(xu ⊗ yu) are Jordan decompositions of x ⊕ y ∈ GL(V ⊕ W ) and x ⊗ y ∈ GL(V ⊗ W ).

Theory of Jordan decompositions generalize to infinite dimensional vector spaces V providing we restrict our attention to locally finite endomorphisms x, i.e. endomorphisms such that any v ∈ V belongs to a finite dimensional x-invariant subspace. A locally finite endomorphism x of V is semisimple if its restriction to every finite dimensional x-invariant subspace of V is semisimple. Nilpotent and unipotent are defined similarly. For a general locally finite x ∈ End V we have its Jordan decompositions x = xs + xn and x = xsxu, with all the properties of the finite dimensional case holding. To define xs, take v ∈ V , find a finite dimensional x-invariant subspace W containing v and define xs(v) = (x|W )s(v). The fact that this is well-defined follows from the uniqueness statement in the finite dimensional Jordan decomposition. The elements xn and xu are defined similarly.

Theorem 11.1.4 For any x ∈ G, there are unique elements xs, xu ∈ G such that (ρx)s = ρxs , (ρx)u = ρxu , and x = xsxu = xuxs. Moreover, if ϕ : G → H is a morphism of algebraic groups, then ϕ(xs) = ϕ(x)s and ϕ(xu) = ϕ(x)u.

Proof Let m : k[G] ⊗ k[G] → k[G] be the algebra multiplication. We have

m ◦ (ρx ⊗ ρx) = ρx ◦ m. 11.1 Jordan-Chevalley decomposition 123 Hence by Lemmas 11.1.2(v) and 11.1.3,

m ◦ ((ρx)s ⊗ (ρx)s) = (ρx)s ◦ m, i.e. (ρx)s respects the multiplication on k[G]. Also ρx(1) = 1 implies (ρx)s(1) = 1 by the properties of Jordan decomposition. Thus (ρx)s is an automorphism of k[G]. Hence ξ : f 7→ ((ρx)sf)(e) is an algebra homomorphism k[G] → k. So there is a point xs ∈ G with ξ(f) = f(xs).

To prove that (ρx)s and ρxs are the same note that λy and ρx commute for all y, so λy and (ρx)s commute too. Now,

((ρx)sf)(y) = (λy−1 (ρx)sf)(e) = ((ρx)sλy−1 f)(e)

−1 = (λy f)(xs) = f(yxs) = (ρxs f)(y).

Similarly we ﬁnd xu such that (ρx)u = ρxu . But the right regular representation is faithful, so ρx = ρxs ρxu = ρxu ρxs implies x = xsxu = xuxs. ∗ ∗ Now, let x ∈ G and y = ϕ(x). It is easy to check that ϕ ◦ρy = ρx ◦ϕ . ∗ ∗ ∗ ∗ Hence ϕ ◦ (ρy)s = (ρx)s ◦ ϕ . So ϕ ◦ ρys = ρxs ◦ ϕ . For any f ∈ k[H],

∗ (ϕ (ρys (f)))(e) = (ρys (f))(ϕ(e)) = (ρys (f))(e) = f(ys). This equals

∗ ∗ (ρxs ◦ ϕ )(f)(e) = (ϕ f)(xs) = f(ϕ(xs)).

We conclude that ϕ(xs) = ys. The argument for the unipotent parts is similar.

Remark 11.1.5 One can also prove the inﬁnitesimal analogue of this result: for any X ∈ g, there are unique elements Xs,Xn ∈ g such that (∗X)s = ∗Xs,(∗X)u = ∗Xn,[Xs,Xn] = 0, and X = Xs +Xn; moreover, if ϕ : G → H is a morphism of algebraic groups, then dϕ(Xs) = dϕ(X)s and dϕ(Xn) = dϕ(X)n. See Humphreys for details.

Decompositions x = xsxu and X = Xs +Xn coming from the theorem and the remark are refereed to as the abstract Jordan decompositions or Jordan-Chevalley decompositions. If x = xs, we call x semisimple, and of x = xu we call u unipotent. The set of all semisimple (resp. unipotent) elements of G is denoted Gs (resp. Gu).

Example 11.1.6 If x ∈ G = GLn, then xs is just the semisimple part m of x considered as an endomorphism of V = k , and xu is the unipotent part. To see this, let f ∈ V ∗ be a non-zero functional. For v ∈ V define 124 Semisimple and unipotent elements f˜(v) ∈ k[G] by f˜(v)(x) = f(xv). This gives an injective linear map ˜ ˜ ˜ f : V → k[G] which satisfies f(xv) = ρxf(v). Hence ˜ ˜ ˜ ˜ f(xsv) = (ρx)sf(v) = ρxs f(v) = f(xsv), where the first xs is the semisimple part of x in the old sense of linear algebra, and the other two xs’s refer to the semisimple part of x in the abstract Jordan decomposition. This implies that the two are the same. The argument for the unipotent parts and Lie algebras is similar. For an arbitrary G, we can embed it as a closed subgroup of some GL(V ). Then again, the abstract Jordan decompositions x = xsxu of x as an element of G and as an endomorphism of V coincide.

11.2 Unipotent algebraic groups An algebraic group is called unipotent if all of its elements are unipotent.

Theorem 11.2.1 Let G be a unipotent closed subgroup of GLn. Then −1 there is g ∈ GLn such that gGg < Un.

Proof Let V = kn. It suffices to show that G fixes some flag in V . Using induction on n we may assume that G does not stabilize any subspace of V , i.e. G acts irreducibly on G. Then by Wedderburn theorem the elements of G span the vector space End V . Since G is unipotent, all elements of G have trace n. Hence 0 = tr(h − gh) = tr(1 − g)h for all g, h ∈ G, hence for all g ∈ G and all h ∈ End V . Taking h to be various matrix units, you now get that 1 − g = 0, i.e. G = {e}.

Corollary 11.2.2 Unipotent algebraic groups are nilpotent.

Theorem 11.2.3 (Rosenlicht) Let G be an unipotent algebraic group acting on an algebraic variety X. Then all orbits of G on X are closed.

Proof Let O be an orbit. Replacing X by O¯, we may assume that O is open dense in X. Let Y be its complement. Consider the action of G on k[X] by translation of functions. This action is locally finite, see Problem 8.5.9. Moreover, G stabilizes Y , so it leaves I(Y ) invariant. By Theorem 11.2.1, there is a non-zero function f ∈ I(Y ) fixed by G. But then f is constant on O. So, since O is dense, f is constant on X. This shows that f is a non-zero scalar, hence I(Y ) = k[X] and Y = ∅. 11.3 Problems 125 Now let G be an arbitrary connected algebraic group. Suppose that X,Y are two closed connected normal solvable subgroups of G. Then XY is again a closed connected normal solvable subgroup of G. It follows that G contains a unique maximal closed connected normal solvable subgroup. This is called the radical of G and denoted R(G). Similarly one defines the unipotent radical Ru(G) as the unique maximal closed connected normal unipotent subgroup. A connected algebraic group is called semisimple if R(G) = {e} and reductive if Ru(G) = {e}. Unipotent groups are nilpotent, so semisimple groups are reductive. There is a beautiful structure theory and classification of reductive groups.

Lemma 11.2.4 If M ⊂ Mn(k) is a commuting set of matrices, then M is triagonalizable. If M consists of the diagonalizable matrices, then M is diagonalizable.

Proof Linear algebra. See Humpreys, 15.4.

Theorem 11.2.5 Let G be a commutative algebraic group. Then Gs and Gu are closed subgroups of G, connected if G is, and the product map ϕ : Gs × Gu → G is an isomorphism of algebraic groups.

Proof That Gs,Gu are subgroups follows from Lemma 11.1.2(iv). That Gu is closed is Problem 11.3.1. Moreover, Theorem 11.1.4 implies that ϕ is an isomorphism of abstract groups. Now embed G into some GLn. Lemma 11.2.4 allows us to assume that G is a group of upper triangular matrices and that Gs is a group of diagonal matrices. This implies that Gs is also closed. It has to be shown that the inverse map is a morphism or that the maps x 7→ xs and x 7→ xu are morphisms. The second is if the ﬁrst is, −1 as xu = xs x. Now xs is just the diagonal part of the matrix x (why?), so x 7→ xs is a morphism. Now the connectedness of G also implies that of Gs and Gu.

11.3 Problems

Problem 11.3.1 The set of all unipotent elements of G is closed. 126 Semisimple and unipotent elements Problem 11.3.2 Let B be a ﬁnite dimensional k-algebra. If x ∈ AutB, then xs, xu ∈ AutB.

Problem 11.3.3 Let char k = 0. An element of GLn having ﬁnite order must be semisimple.

Problem 11.3.4 Let G be a connected algebraic group of positive dimension. Prove that R(G) = {e} if and only if G has no closed connected commutative normal subgroup. (Hint: see Example 8.2.10).

Problem 11.3.5 If char k = 0, then every unipotent subgroup of GLn is connected.

Problem 11.3.6 If char k = 0 then 1-dimensional unipotent group is isomorphic to Ga. 12 Characteristic 0 theory

Throughout this chapter we assume char k = 0.

12.1 Correspondence between groups and Lie algebras

Theorem 12.1.1 (i) If ϕ : G → G0 is a morphism of algebraic groups then ker dϕ = L(ker ϕ). (ii) If A, B < G are closed subgroups then a ∩ b = L(A ∩ B).

Proof (i) We may assume that ϕ is surjective. Of course, L(ker ϕ) ⊂ ker dϕ. Since ϕ is separable, dϕ is surjective, and the result follows by dimensions.

(ii) Let π : G → G/B be the canonical morphism, so ker dπe = b. Let π0 : A → π(A) be the restriction of π. The ﬁbers of π0 are the cosets of A ∩ B in A, and π0 is separable. (Also π(A) is a variety because it is an A-orbit in G/B). Therefore π(A) =∼ A/(A ∩ B), and 0 now as in (i) we deduce that ker dπe = L(A ∩ B). On the other hand, 0 ker dπe = a ∩ ker dπe = a ∩ b.

Lemma 12.1.2 Let G be connected, ρ : G → GL(V ) be a rational representation and dρ : g → gl(V ) be the corresponding representation of g. Then G and g leave the same subspaces (resp. vectors) invariant.

Proof In view of Theorem 12.1.1(i), we may assume that G < GL(V ). By Problem 9.6.1, L(GL(V )W ) = gl(V )W , and GW = G ∩ GL(V )W , gW = g ∩ gl(V )W . By Theorem 12.1.1(ii), L(GW ) = gW . Finally, G

127 128 Characteristic 0 theory stabilizes W if and only if GW = G and g stabilizes W if and only if gW = g.

Corollary 12.1.3 Let B be a ﬁnite dimensional k-algebra. Then

L(AutB) =∼ Der B.

Proof The proof of Corollary 9.5.2 shows that x ∈ GL(B) is an automorphism if and only if it ﬁxes certain tensor t ∈ B∗⊗B∗⊗B, while X ∈ gl(B) is a derivation if and only if it kills t. Now apply Lemma 12.1.2.

Deﬁnition 12.1.4 Let g = L(G). A Lie subalgebra h of g is called algebraic if h = L(H) for a closed connected subgroup H < G.

Even in characteristic 0 not all subalgebras are algebraic.

Theorem 12.1.5 Assume that G is connected. Then the map H 7→ h is a one-to-one inclusion preserving correspondence between the closed connected subgroups of G and the algebraic Lie subalgebras. Moreover, normal subgroups correspond to ideals.

Proof Suppose L(H) = L(K). Using Theorem 12.1.1(ii), we have L(H ∩ K) = L(H)∩L(K) = L(H). So dim H∩K = dim H, whence H∩K = H. Similarly, H ∩ K = K. It follows that H = K. We already know that h is an ideal if H is normal, see Lemma 9.4.1. Conversely, suppose h ⊂ g is an ideal. Then g stabilizes h via ad, hence G stabilizes h via Ad, see Lemma 12.1.2. But for x ∈ G, Ad x : h → g is the diﬀerential of Int x : H → G. By separability, h = Ad x(h) = L(Int x(H)) = L(xHx−1). Now, by the previous paragraph, H = xHx−1, as they have the same Lie algebra.

Theorem 12.1.6 Let G be a connected algebraic group.

(i) If x ∈ G, then L(CG(x)) = cg(x). (ii) ker Ad = Z(G), and L(Z(G)) = z(g).

Proof (i) Lemma 9.4.4 shows that this is true when G = GLn. In general, embed G as a closed subgroup of some GLn and use Theo- rem 12.1.1(ii). (ii) By Theorem 12.1.1, L(ker Ad) = ker ad = z(g). As Ad = d Int, Z(G) ⊂ ker Ad. Conversely, if x ∈ ker Ad, then g = cg(x) = L(CG(x)), 12.2 Semisimple groups and Lie algebras 129 whence CG(x) = G since they have the same Lie algebras. Thus x ∈ Z(G).

Corollary 12.1.7 A connected algebraic group is commutative if and only if its Lie algebra is abelian.

12.2 Semisimple groups and Lie algebras A Lie algebra (of positive dimension) is semisimple if it does not have non-trivial solvable ideals. This is equivalent to the requirement that the Lie algebra does not have non-zero commutative ideals. Similarly, a connected algebraic group of positive dimension is semisimple if and only if it has no closed connected commutative normal subgroup except {e}, see Problem 11.3.4.

Theorem 12.2.1 A connected algebraic group is semisimple if and only if its Lie algebra is semisimple.

Proof If N < G is a closed connected commutative normal subgroup then n is a commutative ideal of g, so n = 0 forcing N = {e}. Conversely, ◦ let n ⊂ g be a commutative ideal. Deﬁne H := CG(n) . Then h = cg(n) by Lemma 12.1.2. Since n is an ideal, so is cg(n). Hence H is normal in G. Hence Z := Z(H)◦ is also normal in G. By Theorem 12.1.6(ii), z is the center of h, and therefore includes n. But G is semisimple, so Z = {e}, z = 0. This forces n = 0.

Remark 12.2.2 When G is semisimple, g is semisimple, so z(g) = 0, whence Z(G) is ﬁnite, see Theorem 12.1.6.

Corollary 12.2.3 Rational representations of semisimple algebraic groups are completely reducible.

Proof This follows from the similar fact about Lie algebras (known as Weyl’s complete reducibility theorem) together with Theorem 12.2.1 and Lemma 12.1.2.

Theorem 12.2.4 Let G be semisimple. Then Ad G = (Autg)◦ and ad g = Der g. 130 Characteristic 0 theory Proof That ad g = Der g is a well-known result in Lie algebras that all derivations of a semisimple Lie algebra are inner. On the other hand, Ad G ⊆ Aut(g)◦, so it suffices to observe that their dimensions coincide. Well, this follows from dim Ad G = dim G, and dim g = dim Der(g) = dim Aut(G)◦, see Corollary 12.1.3. The theorem shows that a semisimple group can be recovered from its Lie algebra ”up to a finite center”, and goes a long way towards the classification of semisimple algebraic groups in characteristic 0.

12.3 Problems Recall that char k = 0 in this capter.

Problem 12.3.1 Let G be a connected algebraic group, H < G closed connected subgroup. Prove that L(NG(H)) = ng(h) and L(CG(H)) = cg(h).

Problem 12.3.2 Let G be a connected algebraic group, h a subalgebra of g. Prove that L(CG(h)) = cg(h).

Problem 12.3.3 Prove that SL2 is semisimple. 13 Semisimple Lie algebras

We saw that in characteristic 0 a connected algebraic group is semisimple if and only its Lie algebra is semisimple. Semisimple Lie algebras can be classified, and this gives us a first approximation to the classification of semisimple algebraic groups in characteristic 0. It turns out that the semisimple algebraic group in characteristic 0 is determined up to finite central subgroup by its Lie algebra (and it is easy to keep the finite group under control). It turns out that the classification of semisimple groups is essentially the same in arbitrary characteristic, although this is much more difficult to prove. In this chapter we are going to review semisimple Lie algebras and explain how to a semisimple Lie algebra we can associate an algebraic group in arbitrary characteristic. This is going to be roughly half of the classification.

13.1 Root systems We want to review classification of the finite dimensional semisimple Lie algebras over C. The first step is to introduce the abstract notion of a root system.

Deﬁnition 13.1.1 A root system is a pair (E, Φ) where E is a (real) Euclidean space and Φ is a ﬁnite set of non-zero vectors, called roots, in E such that (i) Φ spans E. (ii) α, cα ∈ Φ implies c = ±1.

(iii) For any root α, Φ is invariant under the reﬂection sα in the hyperplane orthogonal to α, i.e. the automorphism

β 7→ β − (β, α∨)α,

131 132 Semisimple Lie algebras where α∨ := 2α/(α, α). (iv) (α, β∨) ∈ Z for all α, β ∈ Z. Given a root system, the Weyl group W is the subgroup of GL(E) generated by the sα for α ∈ Φ. It is a ﬁnite group, since it acts faithfully on the ﬁnite set Φ.

We let Hα = {β ∈ E | (α, β) = 0} be the hyperplane orthogonal to α. The connected components of [ E \ Hα α∈Φ are called the Weyl chambers. Fix a chamber C, which we will call the fundamental chamber. Then one can show that the map

w 7→ wC is a bijection between W and the set of chambers. The choice of C ﬁxes several other things. We let Φ+ be the set of all α ∈ Φ which are in the same half space as C (by this we mean that (γ, α) > 0 for any γ ∈ C). Then, Φ = Φ+ t (−Φ+). Elements of Φ+ are called positive roots. Next, let

+ Π = {α ∈ Φ | Hα is one of the walls of C}. This is called a base for the root system. One can show that Π is actually a basis for the vector space E, and moreover every element of Φ+ is a non-negative integer linear combination of Π. Elements of Π are called simple roots.

The Weyl group W is actually generated by the sα for α ∈ Π, i.e. by the reﬂections in the walls of the fundamental chamber. This leads to the idea of the length `(w) of w ∈ W , which is deﬁned as the minimal length of an expression w = sα1 . . . sαr where α1, . . . , αr are simple roots. Geometrically, `(w) is the number of hyperplanes separating wC from C.

Let Π = {α1, . . . , α`}. Here ` = dim E is the rank of the root system. The Cartan matrix A = (ai,j)1≤i,j≤` is the matrix with

∨ ai,j = (αi, αj ). Since all the Weyl chambers are conjugate under the action of W , the Cartan matrix is an invariant of the root system (up to simultaneous permutation of rows/columns). Here are some basic properties about this matrix: 13.1 Root systems 133

(C1) ai,i = 2.

(C2) For i 6= j, ai,j ∈ {0, −1, −2, −3}.

(C3) ai,j 6= 0 if and only if aj,i 6= 0.

0 Note (C2) is not obvious. It follows because E = hαi, αji together with Φ0 := Φ ∩ E0 is a root system of rank 2. Rank 2 root systems are easy (and fun) to classify. Their Cartan matrices are exactly the following:

2 0 2 −1 A × A : ,A : , 1 1 0 2 2 −1 2 2 −2 2 −1 B : ,G : . 2 −1 2 2 −3 2

Note if ai,j 6= 0, then

(αi, αi)/(αj, αj) = ai,j/aj,i ∈ {1, 2, 3}, so in this case you can work out the ratio of the lengths of the roots αi, αj to each other from the Cartan matrix. A root system is called indecomposable if it cannot be partitioned E = E1 ⊥ E2, Φ = Φ1 t Φ2 where (Ei, Φi) are root systems. An equivalent property is that we cannot order roots in such a way that the corresponding Cartan matrix has block-diagonal form. Thus, for an indecomposable root system, one can work out the ratio of lengths of any pair of roots to each other from the Cartan matrix, hence one completely recovers the form (., .) on E up to a scalar from the Cartan matrix. One also recovers Φ, since the Cartan matrix contains enough information to compute the reflection sαi for each i = 1, . . . , `, and Φ = W Π. So (with the correct definition of an isomorphism—give it!) an indecomposable root system is completely determined up to isomorphism by its Cartan matrix. A convenient shorthand for Cartan matrices is given by the Dynkin diagram. This is a graph with vertices labelled by α1, . . . , α`. There are ai,jaj,i edges joining vertices αi and αj, with an arrow pointing towards αi if (αi, αi) < (αj, αj) Clearly you can recover the Cartan matrix from the Dynkin diagram given properties (C1)–(C3) above. Now I can state the classification of root systems:

Theorem 13.1.2 The Dynkin diagrams of the indecomposable root systems are as given in Figure 13.1. 134 Semisimple Lie algebras

An ••••••••...

Bn ••••••••< ...

Cn ••••••••> ... •H H ... Dn ¨••••••• •¨

E6 •••••

• E7 ••••••

• E8 •••••••

•

F4 ••••<

G2 ••<

Fig. 13.1. Dynkin diagrams of semisimple Lie algebras

13.2 Semisimple Lie algebras Now we sketch how the semisimple Lie algebras are classiﬁed by the root systems. We need to start with a semisimple Lie algebra and build a root system out of it, and vice versa. So we begin with a ﬁnite dimensional semisimple Lie algebra g over C. Then g possesses a non-degenerate invariant symmetric bilinear form (., .), where invariant here means ([X,Y ],Z) = (X, [Y,Z]) (in fact, the converse is also true). Moreover, if g is simple, there is a unique such form up to a scalar. There is a “canonical” choice of non-degenerate form, the Killing form, but we don’t need that here.

Example 13.2.1 Let us consider sln. The bilinear form (X,Y ) = tr(XY ) is non-degenerate and invariant. Let ei,j be the ij-matrix unit and let h be the diagonal, trace zero matrices. We can decompose M sln = h ⊕ Cei,j. i6=j

A basis for h is given by h1, . . . , hn−1 where hi = ei,i − ei+1,i+1. Let 13.2 Semisimple Lie algebras 135

∗ εi ∈ h be the map sending a diagonal matrix to its ith diagonal entry. Note ε1 + ··· + εn = 0, i.e. the εi’s are not independent. Then for any H ∈ h we have

[H, ei,j] = (εi − εj)(H)ei,j, i.e. ei,j is a simultaneous eigenvector for h. We use the word weight in place of eigenvalue, so ei,j is a vector of weight εi − εj. Now you recall that the root system of type An−1 can be deﬁned as the real vector ∗ subspace of h spanned by ε1, . . . , εn, and the roots are

Φ := {εi − εj | i 6= j}.

A base for Φ is given by by α1, . . . , αn−1 where αi = εi − εi+1. Let us ﬁnally write gα := Cei,j if α = εi − εj, i.e. the weight space of g of weight εi − εj. Then M g = h ⊕ gα. α∈Φ

In other words, you “see” the root system of type An−1 when you decompose g into weight spaces with respect to the diagonal matrices. Final note: the inner product giving the Euclidean space structure is induced by the non-degenerate form defined to start with. Indeed if you compute the matrix (hi, hj) you get back the Cartan matrix of type An−1. This example is more or less how things go in general, when you start with an arbitrary semisimple Lie algebra g, with a non-degenerate invariant form (., .). The first step is to develop in g a theory of Jordan decompositions. This parallels the Jordan decomposition we proved for algebraic groups. You call an element X of g semisimple if the linear map ad X : g → g is diagonalizable, and nilpotent if ad X is nilpotent. The abstract Jordan decomposition shows that any X ∈ g decomposes uniquely as X = Xs + Xn where Xs ∈ g is semisimple and Xn ∈ g is nilpotent, and [Xs,Xn] = 0. What is more, if you have a representation of g, i.e. a Lie algebra homomorphism ρ : g → gln, it is true that ρ(Xs) = ρ(X)s and ρ(Xn) = ρ(X)n, where the semisimple and nilpotent parts on the right hand side are taken just as n × n matrices in gln. Thus, the abstract Jordan decomposition is consistent with all other Jordan decompositions arising from all other representations. In particular, semisimple elements of g map to diagonalizable matrices under any matrix representation of g. For sln, ei,j is nilpotent for i 6= j, and ei,i − ei,j is semisimple. Now you introduce the notion of a maximal toral subalgebra or Cartan 136 Semisimple Lie algebras subalgebra h of g (in general maximal toral subalgebra and Cartan subalgebra are different notions but they agree for semisimple algebras). This is a maximal abelian subalgebra all of whose elements are semisimple. It turns out that in a semisimple Lie algebra, maximal toral subalgebras are non-zero, and they are all conjugate under automorphisms of g. Now fix one – it doesn’t really matter which, since they are all conjugate. Importantly, the restriction of the invariant form (., .) on g to h is still non-degenerate. So we can define a map

h∗ → h

∗ mapping α ∈ h to tα ∈ h, where tα is the unique element satisfying (tα, h) = α(h) for all h ∈ h. Now we can lift the non-degenerate form on ∗ ∗ h to h by deﬁning (α, β) = (tα, tβ). Thus, h now has a non-degenerate symmetric bilinear form on it too. For α ∈ h∗, deﬁne

gα = {X ∈ g | [H,X] = α(H)X for all H ∈ g}.

L ∗ Clearly, g = α∈h∗ gα. Set Φ = {0 6= α ∈ h | gα 6= 0}. Then you get Cartan decomposition of g: M g = h ⊕ gα α∈Φ

(it is not obvious that the right hand side is everything...). It turns out with some work that each of the gα spaces are one-dimensional. Now you can build a root system out of g: we’ve already constructed the set Φ. Let E be the real vector subspace of h∗ spanned by Φ. The restriction of the form on h∗ to E turns out to be real valued only, and makes E into a Euclidean space. Now:

Theorem 13.2.2 The pair (E, Φ) just built out of g (starting from a choice of h) is a root system. Moreover, the resulting map from semisimple Lie algebras to Dynkin diagrams gives a bijection between isomorphism classes of semisimple Lie algebras and Dynkin diagrams. The decomposition of a semisimple Lie algebra as a direct sum of simples corresponds to the decomposition of the Dynkin diagram into indecomposable components.

For example, sln is the simple Lie algebra corresponding to the Dynkin diagram An−1. 13.3 Construction of simple Lie algebras 137 13.3 Construction of simple Lie algebras We now explain how to construct the simply-laced simple Lie algebras. So let (E, Φ) be a root system of type A`,D` or E`, π = {α1, . . . , α`} be a base and Φ+ the corresponding set of positive roots. We may assume that (α, α) = 2 for all α ∈ Φ, as all roots have the same length. Let Q = ZΦ ⊂ E be the root lattice, the free abelian group on basis Π. We construct an asymmetry function

ε : Q × Q → {±1}

such that

(1) ε is bilinear, i.e. ε(α + α0, β) = ε(α, β)ε(α0, β) and ε(α, β + β0) = ε(α, β)ε(α, β0) for all α, α0, β, β0 ∈ Q. (2) ε(α, α) = (−1)(α,α)/2 for all α ∈ Q.

Note (2) implies

(3) ε(α, β)ε(β, α) = (−1)(α,β) for all α, β ∈ Q.

To construct such an ε, it suffices by bilinearity to define it on elements of Π. Choose an orientation of the Dynkin diagram. Then define  1 if αi and αj are not connected,   1 if αi → αj, ε(αi, αj) =  −1 if αi ← αj,  −1 if αi = αj.

∗ Now we can construct g. Let h = C ⊗Z Q = C ⊗R E. Let h be the dual space, and let Hα ∈ h be the element such that β(Hα) = (β, α) for ∗ all β ∈ h . Then, H1,...,H` gives a basis for h, where Hi = Hαi . Now let M g = h ⊕ CEα α∈Φ as a vector space. Deﬁne a multiplication by the formulae

[Hi,Hj] = 0,

[Hi,Eα] = α(Hi)Eα = (αi, α)Eα,

[Eα,E−α] = −Hα,

[Eα,Eβ] = 0 if α + β∈ / ϕ ∪ {0},

[Eα,Eβ] = ε(α, β)Eα+β if α + β ∈ Φ. 138 Semisimple Lie algebras Theorem 13.3.1 g is the simple Lie algebra of type Φ, with maximal toral subalgebra h.

Proof You of course have to check that g is a Lie algebra, which boils down to checking that the Jacobi identity is satisﬁed. This is a case analysis. Having done that, we deﬁne a bilinear form on g by

(Hi,Hj) = (αi, αj), (Hi,Eα) = 0, (Eα,Eβ) = −δα,−β.

You check that this is a non-degenerate invariant bilinear form. More- over, h is a toral subalgebra of g, and since the 0-weight space of h on g is just h itself, it must be maximal. Finally, it is automatic that the corresponding root system is of type Φ. Hence, g is simple of type Φ with maximal toral subalgebra h.

Deﬁnition 13.3.2 Let g be an arbitrary semisimple Lie algebra (not necessarily simply-laced). Let Φ be a root system corresponding to a choice of maximal toral subalgebra h, and let Π = {α1, . . . , α`} be a base for Φ. For α, β ∈ Φ, the α-string through β is the sequence

β − rα,...,β,...,β + qα where r and s are the maximal integers such that all the vectors in the string belong to Φ. It turns out that r and q are equal to 0, 1, 2 or 3 in all cases, and 2 and 3 don’t arise if the root system is simply-laced.

Denote Hα := 2tα/(α, α) and Hi := Hαi .A Chevalley basis for g means a basis

{H1,...,H`} ∪ {Xα | α ∈ Φ} such that

(a) [Hi,Hj] = 0, ∨ (b) [Hi,Xα] = (α, αi )Xα, (c) [Xα,X−α] = Hα, and this is a Z-linear combination of H1,...,H`, (d) If α, β, α + β ∈ Φ and β − rα, . . . , β + qα is the α-string through β, then [Xα,Xβ] = Nα,βXα+β = ±(r + 1)Xα+β. The key thing is that all the structure constants in a Chevalley basis are integers!

Theorem 13.3.3 (Chevalley) Chevalley bases exist. 13.4 Kostant Z-form 139 Proof If Φ is simply-laced, this is easy from the above construction: + − take Xα = Eα if α ∈ Φ and −Eα is α ∈ Φ . Now you easily check this satisﬁes the properties. If Φ is not simply-laced, we need some other construction. For classical Lie algebras that is not too hard: you can write them down just as explicitly as sln. Problem 13.6.4 gives an example of how you do this. Another way is to realize all the non-simply- laced root systems as ﬁxed points of automorphisms of simple-laced ones.

13.4 Kostant Z-form Informally speaking, Chevalley group is constructed from a semisimple Lie algebra g as the group generated by the ‘exponents’ of the form exp(tXα) where Xα is a root element of the Lie algebra and t is a scalar. But there are some problems here. Consider, for example

2 3 exp(Xα) = 1 + Xα + Xα/2! + Xα/3! + ...

What does that mean? We don’t have a toplogy to speak of convergence, so we need to make sure that the sum is finite. Well, this will be achieved 3 if Xα is nilpotent in a certain sense. Further, what does Xα mean? We can’t multiply in a Lie algebra! However we can consider this as an element of the universal enveloping algebra. There is a further problem however. If characteristic is 2 or 3, we can’t make sense of the division by 3!. The solution to this is very clever—we will first divide by 3! and then pass to characteristic p!! More formally, we will consider a Z-form n UZ of the universal enveloping algebra U of g which contains all Xα /n! and then pass to the algebra U = Uk := UZ ⊗Z k, called the hyperalgebra. First, recall the universal enveloping algebra U(g) associated to a Lie algebra g. It is defined by a universal property, but there is also an explicit construction. The all-important PBW theorem shows that we can identify g with a Lie subalgebra of U(g), and moreover if X1,...,XN is a basis for g, then the monomials

r1 rn X1 ...XN give a basis for U(g). One reason U(g) is important is because it allows you to view representations, i.e. Lie algebra homomorphisms ρ : g → gl(V ), as U(g)- modules: 140 Semisimple Lie algebras Lemma 13.4.1 The categories of representations of g and of U(g)- modules are isomorphic.

Proof Let ρ : g → gl(V ) be a representation. Viewing gl(V ) as End (V ), we get induced a unique associative algebra homomorphismρ ˆ : U(g) → End (V ). Using this, we make V into a U(g)-module by u.v =ρ ˆ(u)(v). If you think about it, this gives a functor {representations of g} → {U(g)- modules}. Conversely, given a U(g)-module, deﬁne ρ : g → gl(V ) by ρ(X)(v) := Xv. This deﬁnes an inverse functor.

Now we state the main result about the Kostant Z-form.

Theorem 13.4.2 Let g be a semisimple Lie algebra over C, with Cheval- ley basis {H1,...,H`} ∪ {Xα | α ∈ Φ}. Let UZ be the Z-subalgebra of r U(g) generated by the Xα/r! for all α ∈ Π, r ≥ 1. Then, UZ is free as a Z-module with basis given by all monomials in the rα Hi Xα /|rα!(α ∈ Φ), (i = 1, . . . , `) mi in some ﬁxed order, where mi, rα ≥ 0.

Proof (1) Observe all the “Kostant monomials” form a C-basis for U(g) by the PBW theorem, so they are linearly independent. (r) Hi (2) Observe all Xα and all belong to UZ – by constructing mi them as various commutators of the generators of UZ. Hence all Kostant monomials belong to UZ. (3) Prove that the product of two Kostant monomials can be expanded as a Z-linear combination of other Kostant monomials. Hence they span UZ. This is done by proving various commutation relations.

13.5 Weights and representations Let Q = ZΦ ⊂ h∗ be the root lattice. Let P be the weight lattice, deﬁned as ∗ P = {λ ∈ h | λ(Hi) ∈ Z for all i = 1, . . . , `}.

Thus P is the lattice dual to the lattice ZH1 +···+ZH` in h. Obviously, Q ⊆ P . Moreover, since P and Q are both lattices in h∗, i.e. they are both finitely generated abelian groups that span h∗ over C, the quotient 13.5 Weights and representations 141 P/Q is a finitely generated, torsion abelian group. But that implies P/Q is a finite abelian group. It is called the fundamental group. To get a basis for P (as a free abelian group), one can take the fundamental weights ω1, . . . , ω` defined by

ωi(Hj) = δi,j, i.e. the dual basis to H1,...,H`. We claim that

` X αi = ai,jωj j=1

∨ where ai,j = (αi, αj ) = αi(Hj) is the Cartan integer. To see this, just evaluate both sides on Hj – you get the same thing. Thus, P/Q is the abelian group on generatorsω ¯1,..., ω¯j subject to relations

` X ai,jω¯j = 0. j=1 Considering elementary divisors, you get that |P/Q| = det A, indeed, you get an explicit description of P/Q as an abelian group. These are the orders:

A` : ` + 1

B`,C`,E7 : 2

D` : 4

E6 : 3

E8,F4,G2 : 1

In fact the fundamental group is cyclic in all cases except for D` with ` even, when it is Z/2 × Z/2. This is going to be very important: we now know exactly all possible lattices lying between Q and P . The last important ingredient that we need to construct the Chevalley groups is a little representation theory of semisimple Lie algebras. We are interested here just in the finite dimensional representations of g.A fundamental theorem of Weyl (mentioned before) shows that any finite dimensional representation of g decomposes as a direct sum of irrducible representations, i.e. ones with no proper invariant submodules. So we really only need to discuss the finite dimensional irreducible representations. 142 Semisimple Lie algebras Now, if V is a finite dimensional U(g)-module, we can decompose M V = Vλ λ∈h∗ where

Vλ = {v ∈ V | Hv = λ(H)v for all H ∈ h}. This is the weight space decomposition of V . For example, the Cartan decomposition of g itself is the weight space decomposition of the adjoint representation. We will say that v ∈ V is a high weight vector of high + weight λ if 0 6= v ∈ Vλ and Xαv = 0 for all α ∈ Φ . It is obvious that any non-zero ﬁnite dimensional V possesses such a vector – because XαVλ ⊆ Vλ+α and there are only ﬁnitely many non-zero weight spaces in total. It turns out that the high weight vector of an irreducible representation of g is unique up to a scalar and its weight belongs to the set

+ P = {λ ∈ P | λ(Hi) ≥ 0 for each i = 1, . . . , `}. Conversely, for any element of P +, there is a unique up to isomorphism irreducible representation of that highest weight. Relation to Z-forms is as follows:

Theorem 13.5.1 Let λ ∈ P + and V be the corresponding irreducible highest weight representation. Then, there exists a lattice VZ in V invariant under the action of the Kostant Z-form UZ, such that X VZ = Vµ,Z µ∈h∗ where Vµ,Z = Vµ ∩ VZ. Given any ﬁnite dimensional representation V , we consider its lattice L(V ), which is deﬁned to be the subgroup of P generated by all λ such that Vλ 6= 0. If V is faithful, then Q ⊆ L(V ) ⊆ P . In fact you can get any intermediate lattice arising for suitable choice of V , and the possible lattices are parametrized by the subgroups of the fundamental group.

13.6 Problems

Problem 13.6.1 Write down the explicit construction of the root systems of type A`,B`,C` and D`, and show that the length of the longest 13.6 Problems 143 element w0 of the Weyl group was `(` + 1)/2. (Hint: You need to look it up! There are many good sources, e.g. Humphreys’ “Introduction to Lie algebras and representation theory”, Bourbaki “Groupes et Algebres de Lie”, Kac “Inﬁnite dimensional Lie algebras”, Carter “Finite groups of Lie type”, Helgason “Diﬀerential geometry and symmetric spaces”...)

Problem 13.6.2 Look up or work out the dimensions of the simple Lie algebras of types A`,B`,C` and D`. In particular, check that dim C` is the same as the dimension of the algebraic group Sp2`.

Problem 13.6.3 In the proof of Theorem 13.3.1, go through the details needed to verify that the bilinear form deﬁned is invariant.

Problem 13.6.4 Let V be a (2` + 1)-dimensional complex vector space with an ordered basis e1, . . . , e`, e0, e−`, . . . , e−1. Deﬁne a symmetric bilinear form on V by declaring (ei, ej) = 0 (i 6= −j), (ei, e−i) = 1 (i 6= 0) and (e0, e0) = 2. Let J be the matrix of this bilinear form in the basis, ordering rows and columns as e1, . . . , e`, e0, e−`, . . . , e−1. (i) Compute the matrix J explicitly. (ii) Let g = {X ∈ gl(V ) | XT J + JX = 0} be the Lie algebra so(V ) = so(2` + 1). Viewing elements of g as block matrices in our ordered basis, we can write   A v B   X =  r x s  . C w D

Compute explicitly the conditions that the `×` matrices A, B, C, D, the column vectors v, w, the row vectors r, s and the scalar x must satisfy for X to belong to g. (iii) Let h be the set of all diagonal matrices in g. This is a toral ∗ subalgebra of g. Let i ∈ h be the function

diag(t1, . . . , t`, 0, −t`,..., −t1) 7→ ti,

∗ so that ε1, . . . , ε` form a basis for h . Let

Φ = {±εi ± εj, ±εk | 1 ≤ i < j ≤ `, 1 ≤ k ≤ `}, the root system of type B`. Let Ei,j : V → V denote the linear map 144 Semisimple Lie algebras with Ei,j.vk = δjkvi for all −` ≤ i, j, k ≤ `. For α ∈ Φ, deﬁne

α i − j(i < j) i + j(i < j) i Xα Ei,j − E−j,−i Ej,−i − Ei,−j 2Ei,0 − E0,−i X−α Ej,i − E−i,−j E−i,j − E−j,i E0,i − 2E−i,0 Verify that M g = h ⊕ CXα α∈Φ is the Cartan decomposition of g. (v) Deﬁning H1,...,H` appropriately, check that {H1,...,H`}∪{Xα|α ∈ Φ} is a Chevalley basis for g. 14 The Chevalley construction

To motivate the construction, let’s stick to working over C for a bit. Let g be a semisimple Lie algebra, and let V be a ﬁnite dimensional faithful representation. So the lattice L(V ) is some intermediate lattice between Q and P (e.g. if V is the adjoint representation, L(V ) = Q). Let {Hi} ∪ {Xα} be a Chevalley basis. Since V has only ﬁnitely many weight spaces and the Xα map Vµ into Vµ+α, each Xα acts on V nilpotently. Thus, we can consider the formal series

2 2 exp(cXα) = 1 + cXα + c Xα/2! + ... for any scalar c ∈ C as a well-deﬁned endomorphism of V (the inﬁnite sum terminates...). By familiar properties of exponential series,

exp(cXα) exp(dXα) = exp((c + d)Xα).

In particular, exp(cXα) is invertible with inverse exp(−cXα). Now let G be the subgroup of GL(V ) generated by all exp(cXα) for all c ∈ C and α ∈ Φ. This is the Chevalley group corresponding to g in the representation V . It turns out that (up to isomorphism) G is determined by g and the lattice L(V ). Using Z-forms we can imitate this construction over an arbitrary ﬁeld k. In the case that k is an algebraically closed ﬁeld, G is always a semisimple algebraic group, and in fact all semisimple algebraic groups arise out of this contstruction for some choice of Φ and L(V ). In order to study the structure of G in detail, we construct closed subgroups U, T, B, N of G with explicitly named generators, and prove various relations between these generators. We will show that B = UoT (semidirect product), that T ¡ N, and identify the quotient group N/T with the original Weyl group W .

145 146 The Chevalley construction Finally, we will prove the Bruhat decomposition: [ G = BwB. w∈W Here w ∈ W needs to be interpreted as a ﬁxed coset representative in N. Thus, the (B,B)-double cosets in G are parametrized by the Weyl group W .

14.1 Definition and first properties To every field, every root system Φ and every lattice L such that Q ⊆ L ⊆ P we associate the Chevalley group G = G(k, Φ,L) defined as follows. Let UZ be the Kostant Z-form of the universal enveloping algebra U(g) of the semisimple Lie algebra g of type Φ, and VC be a representation of g with L = L(V ). Pick a UZ-invariant lattice VZ in V as in Theorem 13.5.1. Let V = VZ ⊗Z k. Then for any t ∈ k and α ∈ Φ, 2 2 xα(t) := exp(tXα) = 1 + Xα ⊗ t + (Xα/2!) ⊗ t + ... can be considered as an invertible endomorphism of V . By definition, G is the group generated by all xα(t) for t ∈ k and α ∈ Φ. The proof that the group only depends on L and not on the choice of

VC and VZ ⊂ VC will be skipped. For ﬁxed α ∈ Φ, let Xα be the subgroup of GL(V ) generated by all xα(t) for all t ∈ k. From now on we assume as usual that k is algebraically closed.

Theorem 14.1.1 The group G is a closed connected subgroup of GL(V ).

Proof Note that the map Ga → GL(V ), t 7→ xα(t) is a morphism of algebraic groups, as the exponent stops after finitely many steps and so is ”polynomial” in t. So each Xα is a closed connected subgroup of GL(V ). Now use Corollary 8.2.7. The main goal of this course is to prove that: (a) G is a semisimple algebraic group and (b) every semisimple algebraic group is obtained in this way. You have probably already figured out that we are not going to get there by the end of the term but we will try to get as far as possible... Concerning (a), it is a well known fact (often proved even in the 600 algebra courses) that PSLn(k) is simple as an abstract group (all you need for this is the assumption that k has more than 3 elements if n = 2, 14.1 Definition and first properties 147 but remember that for us k is algebraically closed, so infinite). So if G is of type A, any of its solvable normal subgroups is contained in the center of G, which is finite. As the radical is connected, this proves that the radical of G is trivial. The proof for other types uses elements xα(t) instead of transvections in the usual proof for PSLn, and we will not reproduce it here. 15 Borel subgroups and flag varieties

In the previous chapter, we sketched the construction of the semisimple algebraic groups. It is very explicit and case-free. We will now go back to algebraic geometry and sketch the proof of the Classiﬁcation Theorem. We will see that algebraic geometry, in particular the variety structure on quotient varieties G/H which we haven’t really used yet in a deep way, is a fundamental tool to studying group theory.

15.1 Complete varieties and Borel’s ﬁxed point theorem Recall the notion of the complete variety from chapter 7. We need the following:

Lemma 15.1.1 Let G be an algebraic group acting transitively on varieties X,Y . Let ϕ : X → Y be a bijective, G-equivariant morphism. If Y is complete, then X is complete.

Proof By Remark 7.1.3(ii), we need to show that π2 : X × Z → Z is closed for all affine varieties Z. Since Y is complete, it suffices to prove that ϕ × id : X × Z → Y × Z is closed. By Proposition 5.3.2, there are open subsets U ⊂ X and V ⊂ Y such that ϕ(U) = V and ϕ|U : U → V is a finite morphism. Let R, S, T be the respective affine algebras of U, V, Z. Since R is integral over S, R ⊗ T is integral over S ⊗ T . Therefore ϕ × id : U × Z → V × Z is also a finite morphism. In particular, it is a closed map, see Corollary 5.2.4. Because G acts transitively on X, Y and ϕ is G-equivariant, X (resp. Y ) is covered by finitely many translates of the form xU (resp. xV ) for some x ∈ G. It follows that ϕ × id : X × Z → Y × Z is closed.

148 15.2 Borel subgroups 149 Now we can prove the important

Theorem 15.1.2 (Borel’s ﬁxed point theorem) Let G be a connected solvable algebraic group, and X be a non-empty complete G- variety. Then G has a ﬁxed point on X.

Proof Proceed by induction on dim G, the case G = {1} being trivial. Suppose then that dim G > 0 and let H = G0, which is connected solvable of strictly smaller dimension. By induction, Y = {x ∈ X | Hx = x} is non-empty. By Lemma 8.3.1(iii), Y is closed, hence complete, and G stabilizes Y as H ¡ G. So we may as well replace X by Y to assume that H ⊆ Gx for all x ∈ X. Since G/H is abelian, this implies that each Gx ¢ G. Now choose x so that the orbit G · x is of minimal dimension. Then, G · x is closed hence complete. The map G/Gx → G.x is bijective, so we deduce that G/Gx is complete by the preceeding lemma. But G/Gx is aﬃne as Gx ¢ G. So in fact G/Gx is a point, i.e. G = Gx and x is a ﬁxed point.

Corollary 15.1.3 (Lie-Kolchin theorem) Let G be a connected solvable subgroup of GL(V ). Then G ﬁxes a ﬂag in V .

Proof Let G act on the ﬂag variety F(V ). This is projective, so G has a ﬁxed point.

15.2 Borel subgroups Let G be a connected algebraic group.

Deﬁnition 15.2.1 A Borel subgroup B of G is a maximal closed connected solvable subgroup of G.

Example 15.2.2 (i) If G is a Chevalley group, the subgroup B = TU is a Borel subgroup of G. Any conjugate of B in G will give another such subgroup.

(ii) If G = GLn, the subgroup B of all upper triangular matrices is a maximal closed connected solvable subgroup by the Lie-Kolchin theorem. Hence, it is a Borel subgroup. Any conjugate of this will give 150 Borel subgroups and ﬂag varieties another. Note in this case that the quotient variety G/B is the ﬂag variety, so it is a projective variety in particular.

Theorem 15.2.3 For any connected algebraic group G, let B be a Borel subgroup. Then, G/B is a projective variety, and all other Borel subgroups of G are conjugate to B.

Proof Let S be a a Borel subgroup of maximal dimension. Apply Cheval- ley’s theorem to construct a representation ρ : G → GL(V ) and a 1-space L ⊂ V such that S = StabG(L). By the Lie-Kolchin theorem, S fixes a flag in V/L. Hence S fixes a flag F = (L = L1 ⊂ L2 ⊂ · · · ⊂ Ln = V ) in the flag variety F(V ). Recall this is a projective variety, hence it is complete. By the choice of L, S = StabG(F ). Hence the orbit map induces a bijective morphism G/S → G · F ⊂ F(V ). Take any other flag F 0 ∈ 0 F(V ). Then StabG(F ) is upper triangular in some basis, hence it is solvable, hence its dimension is ≤ dim S. This shows that dim G · F 0 = 0 dim G − dim StabG(F ) ≥ dim G · F . Therefore, G · F is a G-orbit in F(V ) of minimal dimension, hence it is closed. This shows that G · F is also complete, so G/S is complete too. Now G/S is complete and its quasi-projective, hence it is projective. Finally, let B be another Borel subgroup of G. Then B acts on G/S, so by Borel’s fixed point theorem, B has a fixed point gS on G/S. Therefore BgS = gS, i.e. g−1Bg ⊆ S. By maximality, we get that g−1Bg = S and this completes the proof.

Deﬁnition 15.2.4 A parabolic subgroup P of G is any closed subgroup of G such that G/P is a projective (equivalently, complete) variety.

Theorem 15.2.5 Let P be a closed subgroup of G. Then, P is parabolic if and only if it contains a Borel subgroup. In particular, P is a Borel subgroup if and only if P is connected solvable and G/P is a projective variety.

Proof Suppose G/P is projective. Let B be a Borel subgroup of G. It acts on G/P with a ﬁxed point, say BgP = gP . This implies that g−1Bg ⊆ P , i.e. P contains a Borel subgroup. Conversely, suppose P contains a Borel subgroup B. The map G/B → G/P is surjective and G/B is complete. Hence, G/P is complete too. But it is quasi-projective too, so in fact G/P is projective. 15.3 The Bruhat order 151

Example 15.2.6 (i) Let G = GLn. The subgroups of G containing B (upper triangular matrices) are exactly the “step” subgroups, one for each way of writing n = n1 + ··· + ns as a sum of positive integers n−1 n1, . . . , ns. There are 2 such subgroups. (ii) Let G be an arbitrary Chevalley group. Let S be a subset of the simple roots Π, so there are 2` possibilities for S. Let P be the subgroup of G generated by B and all sα for α ∈ S. (Equivalently, P is + the subgroup generated by T and the Xα’s for α ∈ Φ ∪ (−S).) Then, P contains B so it is a parabolic subgroup by the theorem, and G/P is a projective variety. In fact, these P ’s give all the parabolic subgroups of G containing the ﬁxed choice of Borel subgroup B. By the theorem, all other parabolic subgroups of G are conjugate to one of these. There are exactly 2` diﬀerent conjugacy classes of parabolic subgroup in the Chevalley group G.

15.3 The Bruhat order

Let G be a Chevalley group, with all the subgroups U, T, Xα,B,N,W = N/T ,. . . . Recall also that W is generated by the simple reflections {sα | α ∈ Π}. For any w ∈ W , we can write w as a product of simple reflections. The length of w was the length of a shortest such expression, called a reduced expression for W . Let me define a partial order on W as follows. Take w, w0 ∈ W . Let 0 w = s1 . . . sr be a reduced expression for w. Declare that w ≤ w if 0 0 and only if w = si1 . . . sij for some 1 ≤ i1 < ··· < ij ≤ r, i.e. if w is a “subexpression” of w. How do you prove this really is a partial order? How do you even show that it is well-defined, i.e. independent of the choice of the reduced expression of w? One of the ways is to use algebraic geometry! By the Bruhat decomposition, the B-orbits on G/B are parametrized by the Weyl group W , i.e. the orbits are the BwB/B’s. Now, the closure of an orbit is a union of orbits, the ones in the boundary being of strictly smaller dimension. So there is obviously a partial ordering ≤ on the orbits of B on G/B defined by O ≤ O0 if and only if O ⊆ O0. What we are going to prove is that this is exactly the partial ordering defined in the previous paragraph! In other words, the ordering in the previous paragraph IS well-defined because there is a geometrically defined partial ordering that amounts to the combinatorics there. Let’s proceed with some lemmas. 152 Borel subgroups and flag varieties Lemma 15.3.1 Let G be an algebraic group, X a G-variety, H ≤ G a closed subgroup and Y ⊆ X a closed H-stable subvariety of X. If G/H is a complete variety (i.e. if H is a parabolic subgroup of G) then G · Y is closed in X.

Proof Let A = {(g, x) ∈ G × X | g−1x ∈ Y }, which is closed in G × X. Let π : G × X → G/H × X be the quotient map. Recall this is an open map. If (g, x) ∈ A then (gh, x) ∈ A for all h ∈ H, since H stabilizes Y . Hence,

π(A) = G/H × X − π(G × X − A), so π(A) is closed in G/H × X. Since G/H is complete, the projection prX (π(A)) ⊆ X is also closed. This is exactly G · Y .

Lemma 15.3.2 Any product of parabolic subgroups of G containing B is closed in G.

Proof Let P1,...,Pr be parabolic subgroups of G containing B. By induction, P2 ...Pr is closed in G and B-stable. Note P2 ...Pr.B = P1 ...Pr−1 is closed and B-stable. Since P1/B is complete, we get by the lemma that P1P2 ...Pr is closed too.

Theorem 15.3.3 (Chevalley) Let G be a Chevalley group. Fix w ∈ W and a reduced expression w = s1 . . . sr for w as a product of simple reﬂections. Then, [ BwB = Bw0B w0

0 where w runs over all subexpressions si1 . . . sij of s1 . . . sr.

Proof Let w = s1 . . . sr be the ﬁxed reduced expression for w, Let

Pi = hB, sii = B ∪ BsiB, where the last equality comes from the work on Chevalley groups in the previous chapter. We show by induction on r that

[ 0 P1 ...Pr = Bw B w0 15.3 The Bruhat order 153 where the union is taken over all subexpressions w0 of the reduced expression s1 . . . sr. The case r = 1 is already done! Now suppose r > 1. Then by induction,

[ 00 P1 ...Pr = Bw B.(B ∪ Bsr B) w00 00 where w runs over all subexpressions of s1 . . . sr−1. But that equals [ Bw0 B w0 00 00 00 00 as required, since Bw srB ⊆ Bw BBsr B ⊆ Bw B ∪ Bw srB by the previous chapter. By the preceeding lemma, P1 ...Pr is closed, hence [ Bw0B, w0 0 union over all subexpressions w of s1 . . . sr, is closed. So it certainly contains the closure BwB. Finally, we know dim BwB is equal to dim B+ the number of positive roots sent to negative roots by w. So in fact we must have that [ Bw0B = BwB w0 by dimension. Now since BwB is defined intrinsically independent of any choice of reduced expression of w, the relation w0 ≤ w iff w0 is a subexpression of some fixed reduced expression for w is well-defined independent of the choice. Moreover, it is a partial ordering on W called the Bruhat ordering. For w ∈ W , the Schubert variety

Xw := BwB/B is a closed subvariety of the flag variety G/B. Note Xw is no longer an orbit of an algebraic group, so it needn’t be smooth. Schubert varieties are extremely interesting projective varieties with many wonderful properties. The Schubert variety Xw0 is the flag variety itself, the Schubert variety X1 is a point. We have shown above that in general, the lattice of containments of Schubert varieties is isomorphic to the Bruhat order on W . 16 The classification of reductive algebraic groups

16.1 Maximal tori and the root system Now we sketch the procedure to build a root system starting from an arbitrary reductive algebraic group. This is the ﬁrst step in proving the classiciation of reductive algebraic groups. Let us start by talking about tori. Recall an n-dimensional torus is an algebraic group isomorphic to Gm ×· · ·×Gm. For example, the subgroup Dn of GLn consisting of all diagonal matrices is an n dimensional torus. Let T be an n-dimensional torus. The character group

∼ ⊕n ∼ n X(T ) = Hom(T, Gm) = Hom(Gm, Gm) = Z . An important point is that, given any two tori T and T 0,

Hom(T,T 0) =∼ Hom(X(T 0),X(T )).

So any homomorphism f : X(T 0) → X(T ) of abelian groups induces a unique morphism T → T 0 of algebraic groups, and vice versa. In fact, you can view X(?) as a contravariant equivalence of categories between the category of tori and the category of ﬁnitely generated free abelian groups. All elements of a torus T are semisimple. So if V is any ﬁnite dimensional representation of T , every element of T is diagonalizable in its action on V by the Jordan decomposition. Moreover, they commute, hence we can actually diagonalize M V = Vλ λ∈X(T ) where

Vλ = {v ∈ V | tv = λ(t)v for all t ∈ T }.

154 16.1 Maximal tori and the root system 155

As before, the Vλ’s are called the weight spaces of V with respect to the torus T . Now let G be an arbitrary connected algebraic group. A maximal torus of G is what you’d think: a closed subgroup T that is maximal subject to being a torus. Let me state some theorems about maximal tori in connected solvable groups. These are proved by induction, though it is often quite diﬃcult...

Theorem 16.1.1 Let G be a connected solvable group. Then, the set Gu of all unipotent elements of G is a closed connected normal subgroup of G. All the maximal tori of G are conjugate, and if T is any one of them, then G is the semi-direct product of T acting on Gu.

As a consequence, you show that in an arbitrary connected group G, all its maximal tori are conjugate. Indeed, any maximal torus T of G is contained in a Borel subgroup B. If T 0 is another maximal torus, contained in a Borel B0, we can conjugate B0 to B to assume that T 0 is also contained in B. But then T and T 0 are conjugate in B by the theorem. Now start to assume that G is a reductive algebraic group. Let T be a maximal torus. Let g be the Lie algebra of G. We can view g as a representation of T via the adjoint action. It turns out moreover – using for the ﬁrst time that G is reductive – that the zero weight space of g with respect to T is exactly the Lie algebra t of T itself. So we can decompose M g = t ⊕ gα α∈Φ where Φ is the set of all 0 6= α ∈ X(T ) such that the T -weight space gα 6= 0. You can already see the root system emerging... The diﬀerence now however is that the set Φ of roots is a subset of the free abelian group X(T ). Now using the assumption that G is reductive again, you show:

(1) Each gα is one dimensional, and α ∈ Φ iﬀ −α ∈ Φ. (2) The group W = NG(T )/T is a ﬁnite group that acts naturally on X(T ) and permutes the subset Φ ⊆ X(T ). (3) Let Q be the root lattice, the subgroup of X(T ) generated by Φ, and

let E = R ⊗Z Q. Fix a positive definite inner product on E that is invariant under the action of W . Then, (E, Φ) is an abstract root system. 156 The classification of reductive algebraic groups (4) If we embed T into a Borel subgroup B, we get a choice Φ+ of positive + + roots defined by α ∈ Φ iff gα ⊂ b. Conversely, any choice Φ of positive roots determines a unique Borel subgroup of G containing T .

We’ve now built out of G a root system (E, Φ), and realized the Weyl group W explicitly as the quotient group NG(T )/T . Moreover, Φ is a subset of the character group X(T ) of T . If G is semisimple, then G is determined up to isomorphism by its root system (E, Φ) together with the extra information given by the fundamental group X(T )/Q. In the next section, we will see a more natural setup which classiﬁes the reductive, not just semisimple, groups. This is harder, since X(T ) will in general be of bigger rank than Q, and so there is much more freedom not captured by the fundamental group alone. For GLn, X(T ) is a free abelian group of rank n, whereas Q is of rank (n − 1).

16.2 Sketch of the classification Finally let’s prepare the way to state the classification of reductive algebraic groups in general. Let G be a reductive algebraic group, and let T be a maximal torus. Let Φ ⊂ X(T ) be the root system of G, defined from the decomposition M g = t ⊕ gα. α∈Φ Let

X(T ) = Hom(T, Gm) be the character group of T , and let

Y (T ) = Hom(Gm,T ) be the cocharacter group. This is also a free abelian group of rank dim T . Moreover, there is a pairing

X(T ) × Y (T ) → Z deﬁned as follows. Given λ ∈ X(T ) and ϕ ∈ Y (T ), the composite λ ◦ ϕ is a map Gm → Gm. So since Aut(Gm) = Z, (λ ◦ ϕ)(x) = xhλ,ϕi

for a unique hλ, ϕi ∈ Z. 16.2 Sketch of the classiﬁcation 157 For each α ∈ Φ, you prove that there is a (unique up to scalars) homomorphism

xα : Ga → G such that −1 txα(c)t = xα(α(t)c)

for all c ∈ Ga, t ∈ T , such that the tangent map

dxα : L(Ga) → gα

is an isomorphism. Moreover, the xα’s can be normalized so that there is a homomorphism

ϕα : SL2 → G such that 1 c 1 0 ϕ = x (c), ϕ = x (c). α 0 1 α α c 1 −α Deﬁne c 0 α∨ : → T, α∨(c) = ϕ . Gm α 0 c−1 So α∨ ∈ Y (T ). This is called the coroot associated to the root α ∈ Φ. Now we have built a datum (X(T ), Φ,Y (T ), Φ∨), where Φ∨ is the set of all coroots. This is the root datum of G with respect to the torus T . (Actually, since all maximal tori in G are conjugate, it doesn’t depend up to isomorphism on the choice of T .) The notion of root datum is the appropriate generalization of root system to take care of arbitrary reductive algebraic groups, not just the semisimple ones. Here is an axiomatic formulation of the notion of root datum: a root datum is a quadruple (X, Φ,Y, Φ∨) where

(a) X (“characters”) and Y (“cocharacters”) are free abelian groups of finite rank, in duality by a pairing h., .i : X × Y → Z; (b) Φ ⊂ X (“roots”) and Φ∨ ⊂ Y (“coroots”) are finite subsets, and there is a given bijection α 7→ α∨ from Φ to Φ∨. To record the additional axioms, define for α ∈ Φ the endomorphisms ∨ sα, sα of X,Y respectively by ∨ ∨ ∨ sα(x) = x − hx, α iα, sα(y) = y − hα, yiα . Then we have the axioms: 158 The classification of reductive algebraic groups (RD1) For α ∈ Φ, hα, α∨i = 2. ∨ ∨ ∨ (RD2) For α ∈ Φ, sαΦ = Φ, sαΦ = Φ . The datum (X(T ), Φ,Y (T ), Φ∨) built from our algebraic group G earlier is such a gadget. There is a notion of morphism of root datum (X, Φ,Y, Φ∨) → (X0, Φ0,Y 0, (Φ0)∨): a map f : X0 → X that maps Φ0 bijectively onto Φ and such that the dual map f ∨ : Y → Y 0 maps f(α)∨ to α∨ for all α ∈ Φ0. Hence there is a notion of isomorphism of root datums. Now suppose that G, G0 are reductive algebraic groups with maximal tori T,T 0 respectively and corresponding root data (X(T ), Φ,Y (T ), Φ∨) and the primed version. Let f :(X(T ),... ) → (X0(T ),... ) be a morphism of root data. It induces a dual map f : T → T 0 of tori. The step is to show that f can be extended to a homomorphism f¯ : G → G0. Using it you prove in particular the isomorphism theorem:

Theorem 16.2.1 Two reductive algebraic groups G, G0 are isomorphic if and only if their root datums (relative to some maximal tori) are isomorphic. There is also an existence theorem:

Theorem 16.2.2 For every root datum, there exists a corresponding reductive algebraic group G. Finally, one intriguing thing: given a root datum (X, Φ,Y, Φ∨) there is the dual root datum (Y, Φ∨,X, Φ). If G is a reductive algebraic group with root datum (X, Φ,Y, Φ∨) you see there is a dual group G∨ with the corresponding dual root datum. Note the process of going from G to G∨ is very clumsy: I don’t think there is any direct way of constructing the dual group out of the original.

Example 16.2.3 Suppose that G is a semisimple algebraic group. Let Q = ZΦ ⊂ X(T ). Here, Q and X(T ) have the same rank, so Q is a lattice in X(T ), and X(T )/Q is a ﬁnite group, the fundamental group. Let P be the dual lattice to Q. Fixing a positive deﬁnite W -invariant inner

product on E = R⊗Z Q, we can identify P with the weight lattice of the root system of G, and then everything is determined by the relationship between Q ⊆ X(T ) ⊆ P . You can formulate the classiﬁcation just of the semisimple algebraic groups in these terms. 16.2 Sketch of the classiﬁcation 159 Example 16.2.4 Let G be a semisimple algebraic group, and suppose that Q ⊆ X(T ) ⊆ P are as in the previous example. If X(T ) = P , then G is called the simply-connected group of type Φ. If X(T ) = Q, then G is called the adjoint group of this type. Now let Gsc be the simply- connected one, Gad be the adjoint one. Let G be any other semisimple group of type Φ. Then, there is an inclusion X(T ) ,→ P = X(Tsc). This induces a map Gsc G. Similarly, there is always a map G Gad.

Example 16.2.5

(1) Consider the root datum of GL2. Here, X(T ) has basis ε1, ε2, these being the characters picking out the diagonal entries. More- ∨ ∨ over, the positive root is α = ε1 − ε2. Also Y (T ) has basis ε1 , ε2 , the dual basis, mapping Gm into each of the diagonal slots. The ∨ ∨ ∨ coroot is α = ε1 − ε2 . (2) GL2 is its own dual group. (3) Consider the root datum of SL2 × Gm. Here, X(T ) has basis α/2, ε, Y (T ) has the dual basis α∨, ε∨ (here α is the usual positive root of SL2). (4) Consider the root datum of PSL2 × Gm. Here, X(T ) has basis ∨ α, ε, Y (T ) has the dual basis α /2, ε. So PSL2 × Gm is the dual group to SL2 × Gm. (5) As an exercise in applying the classiﬁcation, you can show that (1),(3) and (4) plus one more, the 4 dimensional torus, are all the reductive algebraic groups of dimension 4.

Example 16.2.6 Here are some more examples of dual groups (I think!). The dual group to SLn is PSLn. The dual group to Sp2n is SO2n+1. The dual group to P Sp2n is Spin2n+1. The dual group to SO2n is SO2n. The dual group to Spin2n is PSO2n. For more explicit constructions of root datums, see Springer, 7.4.7. Bibliography

[Br] J. Brundan, Lecture Notes on Algebraic Groups. [Hu] J. Humphreys, Linear Algebraic Groups. [Ma] H. Matsumura, Commutative Algebra. [Sh] I.R. Shafarevich, Basic Algebraic Geometry. [Sp] T.A. Springer, Linear Algebraic Groups. [St] R. Steinberg, Lectures on Chevalley Groups.

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